[{"content":"Amadeus Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 80% XII 80% UG 7.5 GPA Round 1 05/10/23\nThere were 2 sections and of 31 Questions and 90 min were given.\nTechnical MCQ (29) Coding (2) Coding Questions Number of Sale Days: A shop operates for N days, an array of size N is given where each day the value can either be 1, 0, -1.\nIf v = 1, then the shopkeeper restocks his inventory and discard his old one (if any). He can\u0026rsquo;t sell anything on that day.\nIf v = 0, then the shopkeeper can sell the items on that day (if he has any items).\nIf v = -1, then the shopkeeper will return back all the item in the shop back to the inventory and can\u0026rsquo;t sell until he restocks.\nFind the maximum number of days the shop has items on sale for one lot of items. (lot refers to the items he gets upon a restock)\nEx:\n4 1 1 0 0 Output: 2 The shop can be on sale for a max of 2 days on 1 lot. 5 1 0 -1 0 0 Output: 1 The shop has nothing in inventory after 3rd day. 5 1 0 0 1 0 Output: 2 int solve(vector\u0026lt;int\u0026gt;\u0026amp; shop) { int n = shop.size(); bool canSell = false; int res = 0; int cur = 0; for (int x: shop) { if (canSell) { if (x == 0) { ++cur; } else { res = max(res, cur); cur = 0; if (x == -1) { canSell = false; } } } else { if (x == 1) { canSell = true; } } } return res; } Free Intervals: A ground is occupied during certain intervals, return an array of intervals when the grouns is free to use. [0, 1e9] is the range of hours.\nGiven\nN: Number of intervals when ground is in use. L: An array of size N which represents the start time of the intervals. R: An array of size N which represents the end time of the intervals. No two intervals overlap, and R[i] \u0026lt; L[i + 1] for 0 \u0026lt; i \u0026lt; UL.\nIf no intervals exist, return [[-1, -1]]\nvector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt; solve(vector\u0026lt;int\u0026gt; R, vector\u0026lt;int\u0026gt; L, int N) { vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt; res; int end = 1e9; for (int i = 0; i \u0026lt; N; ++i) { if (i == 0) { if (L[i] == 0) { res.push_back({0, L[i]}); } } else if (i == N - 1) { if (R[i] != end) { res.push_back({R[i], end}); } } else { if (R[i - 1] \u0026lt; L[i]) { res.push_back({R[i - 1], L[i]}); } } } if (res.size() \u0026gt; 0) { return res; } return vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt; {{-1, -1}}; } Round 1 05/10/23\nAn SHL assessment of 2 sections.\nAptitude (19) Verbal (25) ","permalink":"https://manasch.github.io/placements/code/amadeus/","summary":"Amadeus Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 80% XII 80% UG 7.5 GPA Round 1 05/10/23\nThere were 2 sections and of 31 Questions and 90 min were given.\nTechnical MCQ (29) Coding (2) Coding Questions Number of Sale Days: A shop operates for N days, an array of size N is given where each day the value can either be 1, 0, -1.","title":""},{"content":"Amazon HackOn 2023 Details Job Status Hackathon. Winners offered internships/FTE and prize money.\nCriteria 6.5 CGPA Engineering students 2-4 members per team Prizes Winner - INR 1,00,000 1st runner-up - INR 75,000 2nd runner-up - INR 50,000 Mentorship for the Top 8 teams SWAG for the Top 8 teams The Titan - Top 50 coders - Exciting Prizes The Wonder Woman - Top 5 Female Coders - Exciting Prizes The Trailblazer - Top 5 Fastest Coders - Exciting Prizes The Zen Master - Top 5 Cleanest Coders - Exciting Prizes Round 1 28/09/2023\nEvery member of team had to solve 20 MCQs in 20 minutes. The questions were based on DSA and coding fundamentals.\nRound 2 (Coding) 29/09/2023\nEvery member of team had to solve 2 coding questions in 90 minutes. Problems were random and of various difficulty levels. As a team of 3, we had to solve 6 questions in total mentioned below.\nGiven n strings, determine the pair with the max value product. n is upto 10^6 and each string has max length of 20. The strings can contain alphanumeric characters. If a string contains any numeric character, it\u0026rsquo;s value is the integer concatenation of all numeric characters. Otherwise, it is the length of the string. Convert all strings to their values and find the two biggest values. Multiply them to get the answer.\nExample:\n3 abc a01 a42 Answer: Values are [3, 1, 42]. Max product is 42 * 3 = 126 n = int(input()) s = input().split() a = [] for i in s: if i.isnumeric(): a.append(int(i)) else: value = 0 has_numeric = False for j in i: if j.isnumeric(): value = value * 10 + int(j) has_numeric = True if not has_numeric: value = len(i) a.append(value) a.sort() print(a[-1] * a[-2]) Find the longest subsequence in an array such that each element is 3 times more than the previous element. n is upto 100 and a[i] is upto 2000. #include \u0026lt;cmath\u0026gt; #include \u0026lt;cstdio\u0026gt; #include \u0026lt;vector\u0026gt; #include \u0026lt;iostream\u0026gt; #include \u0026lt;algorithm\u0026gt; void solve() { int n; std::cin \u0026gt;\u0026gt; n; std::vector\u0026lt;int\u0026gt; a(n); for (int i = 0; i \u0026lt; n; ++i) std::cin \u0026gt;\u0026gt; a[i]; std::vector\u0026lt;int\u0026gt; dp(n + 1); for (int i = 1; i \u0026lt;= n; ++i) { int max = 0; for (int j = 1; j \u0026lt; i; ++j) { if (a[j - 1] * 3 == a[i - 1]) max = std::max(max, dp[j]); } dp[i] = 1 + max; } std::cout \u0026lt;\u0026lt; *std::max_element(dp.begin(), dp.end()) \u0026lt;\u0026lt; std::endl; } int main() { std::ios::sync_with_stdio(false); std::cin.tie(nullptr); solve(); return 0; } Given a number, repeatedly find the sum of digits and replace the number. Keep doing this until the number has only a single digit. Print the product of the original number and the single digit. number = int(input()) temp = number while temp \u0026gt;= 10: sum = 0 while temp \u0026gt; 0: sum += temp % 10 temp //= 10 temp = sum print(number * temp) There are n number of rooms in a house. You are given a string s of length 2n-2 with upper and lower case characters. Lower case characters represent keys and Upper case characters represent rooms. All odd indices (1-based) are keys and even indices are rooms. To unlock a room, you need to have its corresponding key. Count the number of rooms for which you\u0026rsquo;ll have to additionally purchase keys if you start from room 1 and visit all rooms in the order given in the string. Example:\n4 aAbBcC Answer: 0. You can visit all rooms without purchasing any keys. 5 aAcBbCaB Answer: 1. You need to purchase key `b` for room `B` at index 4 (1-based). #include \u0026lt;bits/stdc++.h\u0026gt; void solve() { int n; std::string s; std::cin \u0026gt;\u0026gt; n \u0026gt;\u0026gt; s; std::multiset\u0026lt;char\u0026gt; keys; int ans = 0; for (int i = 0; i \u0026lt; n - 1; ++i) { char key = s[2 * i]; char door = s[2 * i + 1]; keys.insert(key); auto f = keys.find(std::tolower(door)); if (f == keys.end()) ++ans; else keys.erase(f); } std::cout \u0026lt;\u0026lt; ans \u0026lt;\u0026lt; \u0026#39;\\n\u0026#39;; } int main() { std::ios::sync_with_stdio(false); std::cin.tie(nullptr); solve(); return 0; } You are given a string s and integer k. You need to encode the string using the following algorithm: Traverse s from left to right If the character is a vowel, you need to increment k by 2. Otherwise, increment k by 1. The value corresponding to the current character is its ASCII value times k Example:\nabecd 1 Answer: [291, 392, 606, 693, 800] 0: 97 * 3 = 291 1: 98 * 4 = 392 2: 101 * 5 = 606 3: 99 * 6 = 693 4: 100 * 7 = 800 Solution is to simulate the above behaviour.\nGiven n numbers, count the unique numbers subject to the following constraints: A number is unique if it is the maximum number in the collection and has only one occurrence. If maximum number has more than one occurrence, it is not unique and all its occurrences are removed from the collection. If maximum number has only one occurrence, it is unique. The number max is removed from the occurence and floor(max / 2) is added to the collection if it is greater than 0. Example:\n5 1 2 3 4 5 Answer: 3 0: [1, 2, 3, 4, 5], unique = 0 + 1 = 1 (5 is unique) 1: [1, 2, 3, 3, 4], unique = 1 + 1 = 2 (4 is unique) 2: [1, 2, 2, 3, 3], unique = 2 + 0 = 2 (3 is not unique and all occurrences are removed) 3: [1, 2, 2], unique = 2 + 0 = 2 (2 is not unique and all occurrences are removed) 4: [1], unique = 2 + 1 = 3 (1 is unique) 5: [], answer = 3 #include \u0026lt;bits/stdc++.h\u0026gt; int main() { std::ios::sync_with_stdio(false); std::cin.tie(nullptr); int n; std::cin \u0026gt;\u0026gt; n; std::vector\u0026lt;int\u0026gt; v(n); for (int i = 0; i \u0026lt; n; ++i) std::cin \u0026gt;\u0026gt; v[i]; std::map\u0026lt;int, int\u0026gt; map; for (int i: v) ++map[i]; int ans = 0; while (!map.empty()) { auto [k, v] = *map.rbegin(); map.erase(k); if (k / 2 \u0026gt; 0) ++map[k / 2]; if (v == 1) ans += 1; } std::cout \u0026lt;\u0026lt; ans \u0026lt;\u0026lt; \u0026#39;\\n\u0026#39;; return 0; } Round 3 (Ideation) 03/10/2023 - 10/10/2023\nYet to happen.\nRound 4 (Prototype) 12-10-2023 - 24-10-2023\nYet to happen.\nRound 5 (Finale) 03-11-2023 - 05-11-2023\nYet to happen.\n","permalink":"https://manasch.github.io/placements/code/amazon_hackon/","summary":"Amazon HackOn 2023 Details Job Status Hackathon. Winners offered internships/FTE and prize money.\nCriteria 6.5 CGPA Engineering students 2-4 members per team Prizes Winner - INR 1,00,000 1st runner-up - INR 75,000 2nd runner-up - INR 50,000 Mentorship for the Top 8 teams SWAG for the Top 8 teams The Titan - Top 50 coders - Exciting Prizes The Wonder Woman - Top 5 Female Coders - Exciting Prizes The Trailblazer - Top 5 Fastest Coders - Exciting Prizes The Zen Master - Top 5 Cleanest Coders - Exciting Prizes Round 1 28/09/2023","title":""},{"content":"Applied Materials Details Job Status Full Time (Employment only)\nCriteria Study Cutoff X 70% XII 70% UG 7 GPA Round 1 29/08/23\nThere were no coding questions, just aptitude and analytical, math and technical questions\n","permalink":"https://manasch.github.io/placements/code/applied_materials/","summary":"Applied Materials Details Job Status Full Time (Employment only)\nCriteria Study Cutoff X 70% XII 70% UG 7 GPA Round 1 29/08/23\nThere were no coding questions, just aptitude and analytical, math and technical questions","title":""},{"content":"Arista Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 7.5 GPA Round 1 17/08/23\nCoding Questions Distances Between Nodes: Given a list of elements, create a BST out of the list of elements, and every time a new element is added to the BST, find the distance (the number of edges between every pair of nodes in the BST) of every possible pair and print the sum. This but after creating a BST. GFG Circular Linked List: Given a list of elements, create a circular linked list in a sorted order, don\u0026rsquo;t assume the first element to be the smallest, after creating the list. Insert 5 to the linked list and print out the values in ascending order. Remember take input this way while (cin \u0026gt;\u0026gt; t) Decompositions of powers of 2: Given a list of elements, decompose them into powers of 2 after which if some particular power of 2 appears odd number of times among all the decompositions combined, print that value. Take XOR of all values, this way only the bits in binary would be set that appear odd number of times, then print that power. ","permalink":"https://manasch.github.io/placements/code/arista/","summary":"Arista Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 7.5 GPA Round 1 17/08/23\nCoding Questions Distances Between Nodes: Given a list of elements, create a BST out of the list of elements, and every time a new element is added to the BST, find the distance (the number of edges between every pair of nodes in the BST) of every possible pair and print the sum.","title":""},{"content":"ARM Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 70% XII 70% UG 7 GPA Round 1 01/10/23\nThere were no coding questions. 2 Mandatory sections and an optional section, 75 min\nAptitude Technical Digital or Analogue ","permalink":"https://manasch.github.io/placements/code/arm/","summary":"ARM Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 70% XII 70% UG 7 GPA Round 1 01/10/23\nThere were no coding questions. 2 Mandatory sections and an optional section, 75 min\nAptitude Technical Digital or Analogue ","title":""},{"content":"AT\u0026amp;T Details Job Status Internship + Performance Based Conversion\nCriteria Study Cutoff X 80% XII 80% UG 7.5 GPA 8.5+ GPA were selected Round 1 16/11/23\nThere were 3 sections:\nTechnical MCQ (35) - 35 Min - 35 Marks Aptitude MCQ (10) - 10 Min - 10 Marks Coding (5) - 45 Min - 50 Marks Coding Questions Binary Search: Find the target element in a sorted array. Return -1 if not found. int solve(vector\u0026lt;int\u0026gt;\u0026amp; arr, int target) { int n = arr.size(); int beg = 0; int end = n - 1; int mid; while (l \u0026lt;= r) { mid = beg + ((end - beg) \u0026gt;\u0026gt; 1); if (arr[mid] \u0026gt; target) { end = mid - 1; } else if (arr[mid] \u0026lt; target) { beg = mid + 1; } else { return mid; } } return -1; } String Matching: Find out whether the target string exists in the given sentence. bool solve(string s, string p) { return (s.find(p) != string::npos); } int main() { vector\u0026lt;string\u0026gt; inp; string temp; while (cin \u0026gt;\u0026gt; temp) { inp.push_back(temp); } string sentence, word; for (int i = 0; i \u0026lt; inp.size(); ++i) { if (i == inp.size() - 1) { word = inp[i]; } else { sentence += inp[i] + \u0026#34; \u0026#34;; } } cout \u0026lt;\u0026lt; solve(sentence, word); return 0; } Increasing Sum Sequence: The given input is a string of digits, from the 3rd digit onwards, it should be the sum of the previous 2 digits.\nFind out whether the given string is an ISS or not.\nBasically Fibonacci\nLongest Palindromic Subsequence int lcs(string a, string b) { int n = a.size(); vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt; dp(n + 1, vector\u0026lt;int\u0026gt;(n + 1, 0)); for (int i = 1; i \u0026lt;= n; ++i) { for (int j = 1; j \u0026lt;= n; ++j) { if (a[i - 1] == b[j - 1]) { dp[i][j] = 1 + dp[i - 1][j - 1]; } else { dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]); } } } return dp[n][n]; } int solve(string s) { string t = s; reverse(t.begin(), t.end()); return lcs(s, t); } int solve(string s) { int n = s.size(); vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt; dp(n, vector\u0026lt;int\u0026gt;(n, -1)); auto dfs = [\u0026amp;] (auto self, int i, int j) { if (i \u0026lt; 0 || j \u0026gt;= n) { return 0; } if (dp[i][j] != -1) { return dp[i][j]; } if (s[i] == s[j]) { dp[i][j] = (i == j) 1 : 2 + self(self, i - 1, j + 1); } else { dp[i][j] = max(self(self, i - 1, j), self(self, i, j + 1)); } return dp[i][j]; }; int res = 0; for (int i = 0; i \u0026lt; n; ++i) { res = max(dfs(dfs, i, i)); res = max(dfs(dfs, i, i + 1)); } return res; } Longest Continous Increasing Subarray: Given an array, find out the length of the longest increasing subarray. int solve(vector\u0026lt;int\u0026gt;\u0026amp; arr) { int n = arr.size(); int res = 1; int cur = 1; for (int i = 1; i \u0026lt; n; ++i) { if (arr[i] \u0026gt; arr[i - 1]) { ++cur; } else { res = max(res, cur); cur = 1; } } res = max(res, cur); return res; } ","permalink":"https://manasch.github.io/placements/code/att/","summary":"AT\u0026amp;T Details Job Status Internship + Performance Based Conversion\nCriteria Study Cutoff X 80% XII 80% UG 7.5 GPA 8.5+ GPA were selected Round 1 16/11/23\nThere were 3 sections:\nTechnical MCQ (35) - 35 Min - 35 Marks Aptitude MCQ (10) - 10 Min - 10 Marks Coding (5) - 45 Min - 50 Marks Coding Questions Binary Search: Find the target element in a sorted array. Return -1 if not found.","title":""},{"content":"Aurva Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 8 GPA 8 GPA or OpenSource contributor Lunch / Dinner provided at office. Round 1 14/10/23\nThe test was not on any platform, rather a document was provided and we were asked to code it locally and upload it to google drive and share them the link.\nCoding Questions Parking Slots: A parking lot has a number of parking slots. It also maintains a register which contains the entry and exit times of the cars using the parking slots.\nGiven the register entries as input, determine the minimum number of slots that the parking lot should have to accommodate the cars.\nThe entry array and exit array are index based. The 1st index indicates when the 1st car entered and exited and so on.\nNote: if there is an entry and exit at the same time we are counting that as a slot being free.\nExample:\nInput: Entry = [1, 3, 4, 5] Exit = [2, 5, 6, 7] Output: 2 int solve(vector\u0026lt;int\u0026gt;\u0026amp; entry, vector\u0026lt;int\u0026gt;\u0026amp; ex, int n) { if (n == 0) { return 0; } if (n == 1) { return 1; } sort(entry.begin(), entry.end()); sort(ex.begin(), ex.end()); int overlaps = 0, count = 0; int start = 0, end = 0; // Basically have to check the number of overlapping intervals while (start \u0026lt; n \u0026amp;\u0026amp; end \u0026lt; n) { // if the start of the running interval is less than the end of the next one, we need a slot. if (entry[start] \u0026lt; ex[end]) { ++start; ++count; } // otherwise that slot has been used and then we don\u0026#39;t need it anymore. else { ++end; --count; } // at any point in time we keep track of the maximum number of slots required. overlaps = max(overlaps, count); } return overlaps; } Leaf Nodes: A binary tree is given where leaf nodes ar interconnected through left and right pointers, with right leaf nodes linked through right pointers and left leaf nodes connected through left pointers.\nFind out all the leaf nodes that the tree is having.\nExample:\nInput: Pointer to root node TreeNode { int val; TreeNode *left; TreeNode *right; } Output: [3, 4, 5, 5] (Order doesn\u0026#39;t matter) Balanced Subarray: Aurva has n servers lined up in a row. The ith server has the capacity of capacity[i].\nAny application can be deployed on a balanced contiguous subsegment of 3 or more servers. A contiguous segment, [l, r] of servers is said to be balanced if capacity[l] = capacity[r] = sum(capacity[l + 1]...capacity[r - 1]) i.e. the capacity of the servers at the endpoints of the segment should be equal to the sum of the capacity of all the interior servers.\nGiven the capacity of each server in a row, find the number of balanced sub-segments in it.\nExample:\n[(9, 3, 3, 3, 9)] == 2 (9, 3, 3, 3, 9) and (3, 3, 3) sub-array satisfies the criteria [(6, 1, 2, 3, 6)] == 1 (6, 1, 2, 3, 6) satisfies the criteria [(9, 3, 1, 2, 3, 9, 10)] == 2 (9, 3, 1, 2, 3, 9) and (3, 1, 2, 3) satisfies the criteria int solve(vector\u0026lt;int\u0026gt;\u0026amp; arr) { if (arr.size() \u0026lt; 3) { return 0; } int n = arr.size(); vector\u0026lt;int\u0026gt; prefix(n, 0); prefix[0] = arr[0]; // Maintain a prefix sum for finding the sum of a contigious subarray quickly for (int i = 1; i \u0026lt; n; ++i) { prefix[i] = prefix[i - 1] + arr[i]; } unordered_map\u0026lt;int, set\u0026lt;int\u0026gt;\u0026gt; unmap; int res = 0; for (int i = 0; i \u0026lt; n; ++i) { // If the value has been seen before, check if the prefix sum difference in the middle is the same as the end values if (unmap.find(arr[i]) != unmap.end()) { for (int idx: unmap[arr[i]]) { if (i - idx + 1 \u0026gt;= 3 \u0026amp;\u0026amp; prefix[i - 1] - prefix[idx] == arr[i]) { ++res; } } } unmap[arr[i]].insert(i); } return res; } ","permalink":"https://manasch.github.io/placements/code/aurva/","summary":"Aurva Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 8 GPA 8 GPA or OpenSource contributor Lunch / Dinner provided at office. Round 1 14/10/23\nThe test was not on any platform, rather a document was provided and we were asked to code it locally and upload it to google drive and share them the link.\nCoding Questions Parking Slots: A parking lot has a number of parking slots.","title":""},{"content":"Baxter Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 7 GPA Round 1 25/08/23\nThere were multiple sections, aptitude, analytical, technical, electrical but no coding.\n","permalink":"https://manasch.github.io/placements/code/baxter/","summary":"Baxter Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 7 GPA Round 1 25/08/23\nThere were multiple sections, aptitude, analytical, technical, electrical but no coding.","title":""},{"content":"Blue Yonder Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 70% XII 70% UG 7 GPA Round 1 30/10/23\nThere were 4 sections in total, 90 min.\nAptitude (20) - 20 min English (10) - 10 min Technical (15) - 20 min Coding (2) - 40 min Coding Questions Flip Array: Given a 2-d matrix, filled with either \u0026lsquo;R\u0026rsquo; for Red or \u0026lsquo;B\u0026rsquo; for Black, given a set of coordinates, flip the colour at the location. Golden Number: A golden number is defined as the number in which the the largest digit subtracted from the sum of the digits for a number is 0.\nGiven a range [X,Y], find the sum of all the golden numbers.\n","permalink":"https://manasch.github.io/placements/code/blue_yonder/","summary":"Blue Yonder Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 70% XII 70% UG 7 GPA Round 1 30/10/23\nThere were 4 sections in total, 90 min.\nAptitude (20) - 20 min English (10) - 10 min Technical (15) - 20 min Coding (2) - 40 min Coding Questions Flip Array: Given a 2-d matrix, filled with either \u0026lsquo;R\u0026rsquo; for Red or \u0026lsquo;B\u0026rsquo; for Black, given a set of coordinates, flip the colour at the location.","title":""},{"content":"British Telecom Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 - Coding Round 19/09/23\nA total of 13 questions. Time - 120 min.\nMCQ (10) - Technical - (KMP and N-Queens TC was asked) Coding (3) - 1 SQL Coding Questions Maximum Friendship Power: There are N Cities and M Bi-Directional Roads. An array A of size N consists of the number of people in each group (there are N groups). A city is connected to another city if it can be reached by the roads. A set of connected cities form a kingdom.\nGiven these values, generate a permutation array B (of size N) which assigns the groups to the cities respectively such that the maximum friendship power can be formed. Friendship power is determined by the number of unique connections among X people, i.e. (X * (X - 1)) / 2.\nSince this number can be large, give the answer % mod_value (A number was given which I forgot). (Assume 1-indexing).\nExample:\nN = 4\nM = [[1,3], [2,4]]\nA = [6,2,1,3]\nOutput: 39 (2 kingdoms 36 + 3)\nint solve(int n, int m, vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt;\u0026amp; roads, vector\u0026lt;int\u0026gt;\u0026amp; a) { unordered_map\u0026lt;int, vector\u0026lt;int\u0026gt;\u0026gt; adj; for (auto road: roads) { adj[road[0]].push_back(road[1]); adj[road[1]].push_back(road[0]); } priority_queue\u0026lt;int\u0026gt; groups(a.begin(), a.end()); priority_queue\u0026lt;int\u0026gt; kingdoms; unordered_set\u0026lt;int\u0026gt; visited; auto dfs = [\u0026amp;] (auto self, int src, int\u0026amp; count) { if (visited.find(src) != visited.end()) { return; } visited.insert(src); ++count; for (int neigh: adj[src]) { if (visited.find(neigh) == visited.end()) { self(self, neigh, count); } } }; int count; for (int i = 1; i \u0026lt;= n; ++i) { count = 0; dfs(dfs, i, count); if (count \u0026gt; 0) { kingdoms.push(count); } } int mod = 1e9 + 7; // Just taking some random modulus as I don\u0026#39;t remember what they gave. int res = 0; int currentSum = 0; int currentKingdomCount; while (!kingdoms.empty()) { currentKingdomCount = kingdoms.top(); kingdoms.pop(); currentSum = 0; while (currentKingdomCount--) { currentSum += groups.top(); groups.pop(); } res += ((currentSum) * (currentSum - 1) / 2) % mod; } return res; } This passed half the test cases, IDK why :( Number in a Range: Three numbers L, R, K are given, where [L, R] determine a range of numbers to consider. Find the K\u0026rsquo;th number in the range such that the pattern 101 occurs anywhere in the binary representation of the number. If the K\u0026rsquo;th number doesn\u0026rsquo;t exist, return -1. bool containsPattern(int n) { int p = 5; // 5 in binary is 101, the pattern to match. int t = 7; // 7 in binary is 111, when anded with this, will give the group of 3 bits to check. while (p \u0026lt;= n) { if ((n \u0026amp; t) == p) { return true; } p \u0026lt;\u0026lt;= 1; t \u0026lt;\u0026lt;= 1; } return false; } int solve(int l, int r, int k) { int res = -1; for (int i = l; i \u0026lt;= r; ++i) { if (containsPattern(i)) { --k; cout \u0026lt;\u0026lt; i \u0026lt;\u0026lt; \u0026#34; \u0026#34;; } if (k == 0) { res = i; break; } } return res; } This TLE\u0026rsquo;d, some tricky math for optimization maybe Digit DP + Binary Search. Coding Ninjas SQL International and Domestic Flights: Given a table airports consisting of columns (id, code, country) where code is the shortform for a state. Return the total number of international flights and domestic flights.\nExample:\nid code country 1 CAL USA 2 TEX USA 3 FLO USA 4 ALA USA 5 BER GER 6 LUX BEL Output:\nInternational Domestic 9 6 select ( select round(count(*) / 2, 0) from airports a1 join airports a2 where a1.code != a2.code and a1.country != a2.country ) as \u0026#34;International\u0026#34;, ( select round(count(*) / 2, 0) from airports a1 join airports a2 where a1.code != a2.code and a1.country = a2.country ) as \u0026#34;Domestic\u0026#34;; Round 2 - Interview About yourself. Resume based, projects discussions. 3 DSA and 1 Pattern printing Given n, print the pattern as 1, 2, 3 \u0026hellip; n \u0026hellip; 3, 2, 1 Detect a cycle in a linked list. Longest occurances of 1\u0026rsquo;s in a binary array, same but in a circular binary array. Given an array of integers, replace each element with the next greatest element from the array to its right. (Daily Temperatures on LC) ","permalink":"https://manasch.github.io/placements/code/british_telecom/","summary":"British Telecom Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 - Coding Round 19/09/23\nA total of 13 questions. Time - 120 min.\nMCQ (10) - Technical - (KMP and N-Queens TC was asked) Coding (3) - 1 SQL Coding Questions Maximum Friendship Power: There are N Cities and M Bi-Directional Roads. An array A of size N consists of the number of people in each group (there are N groups).","title":""},{"content":"Broadridge Details Job Status Full Time (Employment + Internship Mandatory) - Member Technical\nCriteria Study Cutoff X % XII % UG 7 GPA Round 1 15/09/23\nThere were 6 sections, no coding. Test was for 65 min, not timed section-wise\nCognitive Ability (27) Data Structures (4) Database Concepts (3) Networking (6) OOPS Concepts (4) Operating System (3) ","permalink":"https://manasch.github.io/placements/code/broadridge/","summary":"Broadridge Details Job Status Full Time (Employment + Internship Mandatory) - Member Technical\nCriteria Study Cutoff X % XII % UG 7 GPA Round 1 15/09/23\nThere were 6 sections, no coding. Test was for 65 min, not timed section-wise\nCognitive Ability (27) Data Structures (4) Database Concepts (3) Networking (6) OOPS Concepts (4) Operating System (3) ","title":""},{"content":"Cisco Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 - Coding Round 29/08/23\nA total of 41 questions were asked with 40 of them being MCQ\u0026rsquo;s based on C, Cpp, Java, Aptitude. 1 Coding question\nCoding Questions Snakes and Ladders Leetcode GFG\nRound 2 - Interview Technical based questions. Resume based projects discussion. Real world usage of data structures and algos. Networking. CAP Theorem, examples and real world usage. Streaming, Database Python Round 3 - Managerial Managerial questions. Interests, hobbies, about me. How would you manage a team meeting? How would you connect with teams across the globe and ensure work efficiency? SQL questions, Questions on projects. Round 4 - HR About Cisco and Job role. About Yourself. Flexible timings and job location satisfactory? ","permalink":"https://manasch.github.io/placements/code/cisco/","summary":"Cisco Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 - Coding Round 29/08/23\nA total of 41 questions were asked with 40 of them being MCQ\u0026rsquo;s based on C, Cpp, Java, Aptitude. 1 Coding question\nCoding Questions Snakes and Ladders Leetcode GFG\nRound 2 - Interview Technical based questions. Resume based projects discussion. Real world usage of data structures and algos.","title":""},{"content":"Cloudera Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 8% XII 8% UG 8 GPA Cutoff was increased after initial registration. Round 1 21/08/23\n3 Questions:\nSQL (1) Coding (2) Coding Questions Find the 2nd largest value: Write an SQL statement to print the second largest value from a column, if it doesn\u0026rsquo;t exist, then print null. SELECT DISTINCT PRICE FROM PRODUCTS ORDER BY PRICE DESC LIMIT 1, 1; Should have worked but didn\u0026rsquo;t. Lines passing through special point: Given a 2D plane and a special point (p) and n other points (x1, x2\u0026hellip;xn). Find the number of pairs of points amongst the n points that pass through the special point. Finding the slope of each point wrt to the special point, if for some slope m, there are q number of points with the same slope, then the number of pairs of lines that pass through the special point would be (q * (q - 1))/2\nNearest Prime: Given a list of integers, if the number is not a prime, then find the closest prime and print that number, else do nothing. ","permalink":"https://manasch.github.io/placements/code/cloudera/","summary":"Cloudera Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 8% XII 8% UG 8 GPA Cutoff was increased after initial registration. Round 1 21/08/23\n3 Questions:\nSQL (1) Coding (2) Coding Questions Find the 2nd largest value: Write an SQL statement to print the second largest value from a column, if it doesn\u0026rsquo;t exist, then print null. SELECT DISTINCT PRICE FROM PRODUCTS ORDER BY PRICE DESC LIMIT 1, 1; Should have worked but didn\u0026rsquo;t.","title":""},{"content":"Commvault Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 75% XII 75% UG 80% Round 1 - Coding Round 01/08/23\nTwo job roles were offered, QA and SDE, QA only had python as an option and SDE had an option between JAVA and CPP, I chose CPP.\nThere were 2 sections timed separately.\nCPP Fundamentals and Generics MCQ (13) Coding (3) Coding Questions Reverse alternate K nodes in a Singly Linked List A function to reverse a linked list given head and apply this every other m nodes.\nGiven a BST, and two values p and q, find the largest value in the path from p to q. Both p and q exist. Find the lowest common ancestor and traverse right.\nint solve(Node *head, int p, int q) { Node *p1 = head; while (true) { if (p1-\u0026gt;data \u0026gt; p \u0026amp;\u0026amp; p1-\u0026gt;data \u0026gt; q) { p1 = p1-\u0026gt;left; } else if (p1-\u0026gt;data \u0026lt; p \u0026amp;\u0026amp; p1-\u0026gt;data \u0026lt; q) { p1 = p1-\u0026gt;right; } else { break; } } int maxVal = max(p, q); int result = 0; while (p1 != maxVal) { if (p1-\u0026gt;data \u0026gt; maxVal) { p1 = p1-\u0026gt;left; } else if (p1-\u0026gt;data \u0026lt; maxVal) { p1 = p1-\u0026gt;right; } result = max(result, p1-\u0026gt;data); } return result; } A mountain of q meters high exists, with q-1 supports spaced out every 1 meter. Can jump at most r supports at once. Starting from the bottom find the total number of ways to reach the peak. Have a DP array and keep updating its values from right to left. Similar to Climbing Stairs\nint solve(int q, int r) { vector\u0026lt;int\u0026gt; dp(q); for (int i = q - 1; i \u0026gt;= 0; --i) { int count = 0; for (int j = 1; j \u0026lt; r; ++j) { if (i + j == q) { ++count; } else if (i + j \u0026lt; q) { count += dp[i + j]; } else { break; } } dp[i] = count; } return dp[0]; } Round 2 - Design Round Build a tool to represent two snaps of a filesystem in memory.\nMore detailed document was provided and had to meet certain milestones every few hours.\nThe solution should be highly scalable, and have decent performance.\nSome pre-defined functions with class implementations were provided and we had to complete the function definitions.\n","permalink":"https://manasch.github.io/placements/code/commvault/","summary":"Commvault Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 75% XII 75% UG 80% Round 1 - Coding Round 01/08/23\nTwo job roles were offered, QA and SDE, QA only had python as an option and SDE had an option between JAVA and CPP, I chose CPP.\nThere were 2 sections timed separately.\nCPP Fundamentals and Generics MCQ (13) Coding (3) Coding Questions Reverse alternate K nodes in a Singly Linked List A function to reverse a linked list given head and apply this every other m nodes.","title":""},{"content":"Confluent Details Job Status Internship + PPO\nCriteria Study Cutoff X % XII % UG GPA There were 3 Roles\nData Infrastructure Intern Business Intelligence Intern Project Management Intern Round 1 09/10/23\nData Infrastructure Intern There were 3 coding questions to be solved in 75 min.\nCoding Questions Min Cost: Will update when I remember.\nMax Price: Given a tree which has nodes from 1-n which refer to the version of some product. Each node has a value V associated to it. For each node, find the max product of 2 values in the subtree (The subtree includes the node being considered). If the node is a leaf node, then the max product would be 0.\nReturn a list where list[i] = max product of the subtree for the ith node. (Assume 1-indexing)\nV -\u0026gt; [-10^5, 10^5] N -\u0026gt; [1, ?] Folder Count: An existing folder structure is provided which shows the parent-child relationship between different folders.\n\u0026lsquo;folder-1\u0026rsquo; is always the root.\nThe existing folder structure is given as follows:\nfolder-1 a folder-1 b a c b d folder-1 /\\ a b / \\ c d 3 kinds of queries can be performed\nmkdir a b : Create a new folder b under a. rmdir a : Remove the folder a and all of its subfolders. count a : Return the count of all the subfolders including itself. Given n queries amongst which c of them are the count query, return a list of size c holding the count for the respective folder count function call.\nBusiness Intelligence Intern There were 3 SQL questions and some MCQs, 75 min.\nProject Management Intern A group discussion was held with 8 people on each side and were asked to debate on a topic at hand.\n","permalink":"https://manasch.github.io/placements/code/confluent/","summary":"Confluent Details Job Status Internship + PPO\nCriteria Study Cutoff X % XII % UG GPA There were 3 Roles\nData Infrastructure Intern Business Intelligence Intern Project Management Intern Round 1 09/10/23\nData Infrastructure Intern There were 3 coding questions to be solved in 75 min.\nCoding Questions Min Cost: Will update when I remember.\nMax Price: Given a tree which has nodes from 1-n which refer to the version of some product.","title":""},{"content":"CRED Details Job Status Full Time (Employment + Internship)\nCriteria This was off-campus Round 1 18/11/23\nThere were 3 coding questions to be done in 90 min.\nCoding Questions Unique String: The input is a string consisting of lower case english alphabets. For any character that occurs more than once, it can be deleted.\nGiven a string, delete the characters that occur more than once such that the resultant string consists of all the unique characters from the original string as well as is the lexicographically largest one.\nExample:\nInput: abcba Output: cba Input: aazbbza Output: zba Input: aaccb Output: acb string solve(string input_string) { vector\u0026lt;int\u0026gt; lastOccurance(26, -1); vector\u0026lt;bool\u0026gt; used(26, false); priority_queue\u0026lt;char\u0026gt; letters; int n = input_string.size(); char ch; int idx; for (int i = 0; i \u0026lt; n; ++i) { ch = input_string[i]; idx = ch - \u0026#39;a\u0026#39;; lastOccurance[idx] = i; if (!used[idx]) { used[idx] = true; letters.push(ch); } } used.clear(); used.resize(26, false); string res; for (int i = 0; i \u0026lt; n; ++i) { ch = input_string[i]; idx = ch - \u0026#39;a\u0026#39;; if (letters.top() == ch \u0026amp;\u0026amp; !used[idx]) { res.push_back(ch); letters.pop(); used[idx] = true; } else if (ch == lastOccurance[idx] \u0026amp;\u0026amp; !used[idx]) { res.push_back(ch); used[idx] = true; } while (!pq.empty() \u0026amp;\u0026amp; used[pq.top() - \u0026#39;a\u0026#39;]) { pq.pop(); } } return res; } This did not pass all the TC\u0026rsquo;s, 9/15 Required Strength: You and your friend are playing an online game, this game, initially starts out with some people with a strength value associated to each player. There are N levels to this game and in each new level, one new player gets added (along with the players from the previous level).\nYour friend has a strategy to defeat the Rankth player in each round until the end, and you would like to adopt this strategy as well.\nGiven the initial array of strength, number of levels N, and the Rank, determine the total strength required to achieve this strategy and defeat the player in each level.\nAll the players have unique strength levels.\nExample:\n[3 5 2] --\u0026gt; Initial players strength [4 1 9 8] --\u0026gt; New player that is added at each level Level 1 -\u0026gt; 4 is added to the pool Level 2 -\u0026gt; 1 is added to the pool Level 3 -\u0026gt; 9 is added to the pool Level 4 -\u0026gt; 8 is added to the pool 2 --\u0026gt; Rank Answer: 24 Explanation: Rank 2, means that you will be defeating the 2nd strongest player amongst the existing players in each round. Round 0: [3 5 2] -\u0026gt; 3 Round 1: [3 5 2 4] -\u0026gt; 4 Round 2: [3 5 2 4 1] -\u0026gt; 4 (3 is strength of the 2nd strongest player now) Round 3: [3 5 2 4 1 9] -\u0026gt; 5 Round 4: [3 5 2 4 1 9 8] -\u0026gt; 8 Strength = 3 + 4 + 4 + 5 + 8 = 24 int solve(vector\u0026lt;int\u0026gt;\u0026amp; initial_players, vector\u0026lt;int\u0026gt;\u0026amp; new_players, int rank) { priority_queue\u0026lt;int, vector\u0026lt;int\u0026gt;, std::greater\u0026lt;int\u0026gt;\u0026gt; pq(initial_players.begin(), initial_players.end()); while (pq.size() \u0026gt; rank) { pq.pop(); } int res = pq.top(); for (int i = 0; i \u0026lt; new_players.size(); ++i) { pq.push(new_players[i]); if (pq.size() \u0026gt; rank) { pq.pop(); } res += pq.top(); } return res; } Best Traversal: There\u0026rsquo;s a city with g_nodes number of towns and m undirected roads or connections. The connections are determined by two array of size m, g_from and g_to.\nThe following code is used to generate a list B.\nvoid generateB(vector\u0026lt;int\u0026gt;\u0026amp; a, vector\u0026lt;int\u0026gt;\u0026amp; b) { for (int i = 0; i \u0026lt; a.size(); ++i) { bool found = 0; for (int j = 0; j \u0026lt; b.size(); ++j) { if (a[i] == b[j]) { found = 1; break; } } if (!found) { b.push_back(a[i]); } } } Here, A is the traversal you are required to take such that the list B generated using the above code generates the lexicographically largest list possible from the given set of connections.\nYou are supposed to perform a traversal such that you generate the lexicographically largest list B, you are free to perform whatever traversal you like.\nExample:\nNo. Graph A B 1 5 -\u0026gt; 4 -\u0026gt; 3 -\u0026gt; 2 -\u0026gt; 3 -\u0026gt; 4 -\u0026gt; 1 [5, 4, 3, 2, 1] 2 3 -\u0026gt; 2 -\u0026gt; 3 -\u0026gt; 1 [3, 2, 1] 3 4 -\u0026gt; 3 -\u0026gt; 2 -\u0026gt; 1 [4, 3, 2, 1] 4 5 -\u0026gt; 3 -\u0026gt; 4 -\u0026gt; 3 -\u0026gt; 2 -\u0026gt; 3 -\u0026gt; 4 -\u0026gt; 1 [5, 3, 4, 2, 1] A is the optimial traversal to generate list B with the above code.\nvoid insertToB(vector\u0026lt;int\u0026gt;\u0026amp; a, vector\u0026lt;int\u0026gt;\u0026amp; b, int n) { unordered_set\u0026lt;int\u0026gt; vis; for (int i = 0; i \u0026lt; a.size(); ++i) { if (!vis.count(a[i])) { vis.insert(a[i]); b.push_back(a[i]); } if (b.size() == n) { break; } } } vector\u0026lt;int\u0026gt; bestTraversal(int g_nodes, vector\u0026lt;int\u0026gt;\u0026amp; g_from, vector\u0026lt;int\u0026gt;\u0026amp; g_to) { int m = g_from.size(); unordered_map\u0026lt;int, set\u0026lt;int, std::greater\u0026lt;int\u0026gt;\u0026gt;\u0026gt; adj; for (int i = 0; i \u0026lt; m; ++i) { int src = g_from[i]; int dst = g_to[i]; adj[src].insert(dst); adj[dst].insert(src); } unordered_set\u0026lt;int\u0026gt; vis; vector\u0026lt;int\u0026gt; a, b; priority_queue\u0026lt;pair\u0026lt;int, int\u0026gt;\u0026gt; pq; pq.push({g_nodes, -1}); while (!pq.empty()) { while (vis.count(pq.top().first)) { pq.pop(); } auto [node, parent] = pq.top(); pq.pop(); vis.insert(node); a.push_back(node); if (vis.size() == g_nodes) { break; } for (const int\u0026amp; neigh: adj[node]) { if (node != parent) { pq.push({neigh, node}); } } } insertToB(a, b, g_nodes); return b; } ","permalink":"https://manasch.github.io/placements/code/cred/","summary":"CRED Details Job Status Full Time (Employment + Internship)\nCriteria This was off-campus Round 1 18/11/23\nThere were 3 coding questions to be done in 90 min.\nCoding Questions Unique String: The input is a string consisting of lower case english alphabets. For any character that occurs more than once, it can be deleted.\nGiven a string, delete the characters that occur more than once such that the resultant string consists of all the unique characters from the original string as well as is the lexicographically largest one.","title":""},{"content":"CynLr Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 27/08/23\nThere were 5 sections and we had 3 hours.\nPractical Physics (4) Phsyics MCQ\u0026rsquo;s Aptitude Coding (1) Behavioural Coding Questions Number of triangles: Given a 2d matrix of 1\u0026rsquo;s and 0\u0026rsquo;s, find the total number of equilateral and isosceles triangles that can be formed taking any of the 1\u0026rsquo;s to be the vertices. There were no test cases to be passed, just a text box where code could by typed ","permalink":"https://manasch.github.io/placements/code/cynlr/","summary":"CynLr Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 27/08/23\nThere were 5 sections and we had 3 hours.\nPractical Physics (4) Phsyics MCQ\u0026rsquo;s Aptitude Coding (1) Behavioural Coding Questions Number of triangles: Given a 2d matrix of 1\u0026rsquo;s and 0\u0026rsquo;s, find the total number of equilateral and isosceles triangles that can be formed taking any of the 1\u0026rsquo;s to be the vertices.","title":""},{"content":"Deloitte Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 6.5 GPA Round 1 03/09/23\nThere were 4 sections:\nEnglish Comprehension Aptitude Technical Coding (2) Coding Questions Matching Last Character: Given a list of words and a character to match, print the sorted position (1-index) of the word with the other words if the last letter of the word matches with the character to match. def solve(inputStr, searchCh): words = inputStr.split() words.sort() for i in range(len(words)): if words[i][-1] == searchCh: print(words[i], i + 1) Largest Digit in a Number: Given a number, return the largest digit in the number. ","permalink":"https://manasch.github.io/placements/code/deloitte/","summary":"Deloitte Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 6.5 GPA Round 1 03/09/23\nThere were 4 sections:\nEnglish Comprehension Aptitude Technical Coding (2) Coding Questions Matching Last Character: Given a list of words and a character to match, print the sorted position (1-index) of the word with the other words if the last letter of the word matches with the character to match.","title":""},{"content":"DE Shaw Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 7 GPA Round 1 19/11/23\nThere were 4 sections:\nCoding (20 min) Coding (30 min) Aptitude Technical Totally around 95 min.\nCoding Questions Min-Max Difference: Given a string of characters (only smaller case), find out the number of minimum number of characters to remove such that the difference between the max frequency of chars and the min frequency of chars is m which is given. int solve(string s, int m) { int res = 0; unordered_map\u0026lt;char, int\u0026gt; freq; vector\u0026lt;int\u0026gt; arr; for (char ch: s) { ++freq[ch]; } for (auto\u0026amp; [k, v]: freq) { arr.push_back(v); } sort(arr.begin(), arr.end()); int r = arr.size() - 1; int l = 0; while (l \u0026lt; r \u0026amp;\u0026amp; arr[r] - arr[l] \u0026gt; m) { int a = arr[r - 1] - arr[l]; // Removing the max int b = arr[r] - arr[l + 1]; // Removing the min if (a \u0026lt; b) { res += arr[r]; --r; } else { res += arr[l]; ++l; } } return res; } Greatest String: You are given a string as input (only consisting of smaller case characters), you are also given two empty strings pre and post.\nThe following operations can be performed:\nThe prefix of the input string can be pushed to the end of the pre string. The postfix of the pre string can be pushed to the post string. Using these operations, create the lexicographically string possible.\nExample:\nInput: bcdadb | Input | Pre | Post | |--------|-------|--------| | bcdadb | - | - | | cdadb | b | - | | dadb | bc | - | | adb | bcd | - | | db | bcda | - | | db | bcd | a | | b | bcdd | a | | - | bcddb | a | | - | bcdd | ab | | - | bcd | abd | | - | bc | abdd | | - | b | abddc | | - | - | abddcb | Output: abddcb string solve(string s) { string res; stack\u0026lt;pair\u0026lt;char, int\u0026gt;\u0026gt; st; vector\u0026lt;bool\u0026gt; better(s.size(), false); for (int i = s.size() - 1; i \u0026gt;= 0; --i) { for (int j = i + 1; j \u0026lt; s.size(); ++j) { if (s[j] \u0026lt; s[i]) { better[i] = true; break; } } } for (int i = 0; i \u0026lt; s.size(); ++i) { st.push({s[i], i}); while (!st.empty() \u0026amp;\u0026amp; !better[st.top().second]) { res.push_back(st.top().first); st.pop(); } } while (!st.empty()) { res += st.top().first; st.pop(); } return res; } Round 2 25/11/23\nWas a technical round.\nIntroduce yourself Projects discussions Probabily Puzzle Delete and Earn Given a python code, how would you decide if it is indented properly or not, write code to check if a given python code is indented correctly. Round 3 26/11/23\nWas technical and scenario based.\nInterests, Introduction How does google autofill work, how would you go about implementing such a system. There are millions of records in an excel sheet, and different columns, a particular cell for some employee could be color coded due to certain conditions, similarly another cell could be differently colored due to other conditions. How would you optimize this process in the backend and send the colors to the front end. Ludo game with n players, at max 4 players can play at once, not necessary for all 4 to play a game, find out the minimum number of games required such that each player plays every other player atleast once. Difference between a get and post request, if you give a html page with a button that sends some post request when clicked that could modify the database by mistake and do not want them to make any changes, how would you handle this. ","permalink":"https://manasch.github.io/placements/code/deshaw/","summary":"DE Shaw Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 7 GPA Round 1 19/11/23\nThere were 4 sections:\nCoding (20 min) Coding (30 min) Aptitude Technical Totally around 95 min.\nCoding Questions Min-Max Difference: Given a string of characters (only smaller case), find out the number of minimum number of characters to remove such that the difference between the max frequency of chars and the min frequency of chars is m which is given.","title":""},{"content":"Deutsche Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 7 GPA / 70% Round 1 11/08/23\nThere was only one coding section with 3 questions\nCoding Questions Minimum length of positive sum subarray: Given an array with both negative and positive integers, for each index, find the minimum length subarray such that the sum of all elements in that subarray is positive. Bruteforce approach passed all the test cases, but otherwise could be solved using sliding window.\nInfinite Multiples: Given an array of positive integers, and a value k, find the k\u0026rsquo;th largest number among the multiples of each and every element in the given array. Create a min-heap and maintain its size to be k and add k multiples for each element in the array, this way the minimum element in a min-heap of size k would be the k\u0026rsquo;th largest element.\nMaze Runner: Given a 2-D matrix, each cell takes one of three values.\n\u0026lsquo;.\u0026rsquo; - Indicates an open path\n\u0026lsquo;#\u0026rsquo; - Indicates a door (Obstacle)\n\u0026lsquo;*\u0026rsquo; indicates a wall (Obstacle).\nThe person will start at the top-left cell and has to find the shortest path to the bottom-right cell. They can wear two types of glasses, one which allows them to pass through the door and the other to pass through the wall, only one glass can be worn at once, for both the glasses, find the shortest path to reach the destination. If it can\u0026rsquo;t be reached return -1.\nPerform BFS for both cases, could probably be optimized more with DP, not sure.\n","permalink":"https://manasch.github.io/placements/code/deutsche/","summary":"Deutsche Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 7 GPA / 70% Round 1 11/08/23\nThere was only one coding section with 3 questions\nCoding Questions Minimum length of positive sum subarray: Given an array with both negative and positive integers, for each index, find the minimum length subarray such that the sum of all elements in that subarray is positive.","title":""},{"content":"Dish Networks Technology Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 7.5 GPA Round 1 30/08/23\nNo coding, just multiple sections of technical, aptitude, analytical.\n","permalink":"https://manasch.github.io/placements/code/dish/","summary":"Dish Networks Technology Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 7.5 GPA Round 1 30/08/23\nNo coding, just multiple sections of technical, aptitude, analytical.","title":""},{"content":"Egnyte Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 7.5 GPA Round 1 19/08/23\nThere was only one coding section with 2 questions, everyone had a combination of 2 from a pool of questions.\nCoding Questions Sum of total waiting time: REMOVED\nExamples: 1. 2. 3. 4. Create a queue, push all tasks to it, have a counter that is always increasing. Pop the queue and if after decrementing the value becomes 0, add the current counter val to the total variable with mod, else push it to the back of the queue. Repeat till the queue is empty.\n#include \u0026lt;queue\u0026gt; #include \u0026lt;vector\u0026gt; using namespace std; int solution(vector\u0026lt;int\u0026gt; arr) { queue\u0026lt;int\u0026gt; q; int n = arr.size(); for (int i = 0; i \u0026lt; n; ++i) { q.push(arr[i]); } int hours = 1; int modulo = 1e9; int total = 0; while (!q.empty()) { int task = q.front(); --task; q.pop(); if (task == 0) { total = (total + hours) % modulo; } else { q.push(task); } ++hours; } return total; } Max tiles coverage: ***REMOVED***\n\nint solution(vector\u0026lt;int\u0026gt; \u0026amp;A); \nExamples:\n\n\n\n\nCreate another array of sums of it\u0026rsquo;s adjacent neighbour, perform a 2D House-Robber approach to find the best possible sum of values covered by tiles.\nvoid dfs(vector\u0026lt;int\u0026gt;\u0026amp; tileSum, size_t idx, int\u0026amp; largestSum, int sum, int tiles) { if (tiles == 0) { largestSum = max(largestSum, sum); return; } if (idx \u0026gt;= tileSum.size()) { return; } sum += tileSum[idx]; // cout \u0026lt;\u0026lt; sum \u0026lt;\u0026lt; endl; for (size_t i = idx + 2; i \u0026lt;= tileSum.size() + 2; ++i) { dfs(tileSum, i, largestSum, sum, tiles - 1); } } int solution(vector\u0026lt;int\u0026gt; \u0026amp;A) { // Implement your solution here int n = A.size(); int tiles; if (n \u0026gt;= 6) { tiles = 3; } else if (n \u0026gt;= 4 \u0026amp;\u0026amp; n \u0026lt; 6) { tiles = 2; } else { tiles = 1; } vector\u0026lt;int\u0026gt; tileSum; for (int i = 0; i \u0026lt; n - 1; ++i) { tileSum.push_back(A[i] + A[i + 1]); } int largestSum = 0; for (size_t i = 0; i \u0026lt; tileSum.size(); ++i) { dfs(tileSum, i, largestSum, 0, tiles); } return largestSum; } This would most likely TLE for large test cases. Didn\u0026rsquo;t know how to use 2d DP. int solution(vector\u0026lt;int\u0026gt; arr) { vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt; dp(arr.size() + 2, vector\u0026lt;int\u0026gt; (4, 0)); int max_ele = 0; for(int i = 2; i \u0026lt; dp.size(); i++) { for(int j = 1; j \u0026lt; dp[0].size(); j++) { if(i \u0026gt;= 2 * j) { dp[i][j] = max(dp[i - 2][j - 1] + arr[i - 2], dp[i - 1][j]); if(dp[i][j] \u0026gt; max_ele) { max_ele = dp[i][j]; } } else { dp[i][j] = dp[i - 1][j]; } } } return max_ele; } Bottom-up DP ","permalink":"https://manasch.github.io/placements/code/egnyte/","summary":"Egnyte Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 7.5 GPA Round 1 19/08/23\nThere was only one coding section with 2 questions, everyone had a combination of 2 from a pool of questions.\nCoding Questions Sum of total waiting time: REMOVED\nExamples: 1. 2. 3. 4. Create a queue, push all tasks to it, have a counter that is always increasing.","title":""},{"content":"Eli Lilly Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 14/09/23\nThere were totally 4 sections and it was 2 hours. The sections were not times, could allocate how much ever time required for any section.\nReasoning (24) Programming (8) - Basically all were SQL Quantitative (23) Coding (3) - Everyone got different questions. Coding Questions War Ships: Given two inputs, layers and modulus. Return the total number of ships required mod the modulus.\nEach ship is assigned a value V and the first layer has 1 ship with value 2. For each ship, the number of ships in the next layer are decided by the value V for that ship.\nThe number of ships in the next layer for that ship with value V is\n[V * V + 1] % M\nwhere M is the modulus and the values for those ships will be ranging from [0, ([V * V + 1] % M) - 1].\nExample:\nL = 2, M = 3\nLayer 1: 1 ship with V = 2 (Given)\nLayer 2: 2 ships with V = [0, 1] because [2 * 2 + 1] % 3 = 2.\nOutput would be 0 as total number of ships is 3 and 3 % 3 is 0.\nStealing Window: A thief is trying to steal from a bank. The bank has N guards and each guard will guard the place for a certain interval. The bank will be open for T minutes. The thief needs X minutes to complete the heist, and it can be split, not all the X mins need to be done at once. Given the intervals at which each guard will be in front of the bank. Determine whether it is possible for the thief to perform the heist.\nInput: N, T, X, Intervals = [start, end, start, end, start, end\u0026hellip;]\nFind the available time between the intervals and return \u0026ldquo;Possible Z\u0026rdquo; or \u0026ldquo;Impossible Z\u0026rdquo; where Z is the free time between the intervals that the theif has to perform the heist.\nAron\u0026rsquo;s Gift: Aron shows preference to two characters C1 and C2. He gives a string consisting of just these two characters. Given the string length and the two characters he likes. Return the number of strings (substrings) that are possible such that it starts with C1 and ends with C2.\nInputs:\nN - Length of string C1 - Character 1 C2 - Character 2 S - String Ex:\nN - 3\nC1 - x, C2 - y\nS = xxy\nOutput: 2 - [\u0026ldquo;xxy\u0026rdquo;, \u0026ldquo;xy\u0026rdquo;]\nN - 2\nC1 - a, C2 - b\nS = ab\nOutput: 1 - [\u0026ldquo;ab\u0026rdquo;]\nint solve(int n, char c1, char c2, string s) { int countC1 = 0; int res = 0; for (int i = 0; i \u0026lt; n; ++i) { if (s[i] == c1) { ++countC1; } else { res += countC1; } } return res; } ","permalink":"https://manasch.github.io/placements/code/eli_lilly/","summary":"Eli Lilly Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 14/09/23\nThere were totally 4 sections and it was 2 hours. The sections were not times, could allocate how much ever time required for any section.\nReasoning (24) Programming (8) - Basically all were SQL Quantitative (23) Coding (3) - Everyone got different questions. Coding Questions War Ships: Given two inputs, layers and modulus.","title":""},{"content":"Endor Labs Details Job Status Internship Only (With possible conversion based on performance)\nCriteria Study Cutoff X % XII % UG 8 GPA Round 1 04/11/23\nCoding Questions Log Files: A bank has transaction logs in the form of \u0026ldquo;sender_user_id receiver_user_id amount\u0026rdquo;. Given multiple such logs in an array and a threshold value of transactions, return a list of the user_id\u0026rsquo;s whose total transaction count is greater than the threshold in ascending order.\nExample:\n88 99 200 99 20 100 20 99 200 12 12 200 Threshold: 2 | ID | Transactions | |----|--------------| | 88 | 1 | | 99 | 3 | | 20 | 2 | | 12 | 1 | \u0026gt; 12 is considered as just 1 transaction as it is a self-transfer Output: [\u0026#39;20\u0026#39;, \u0026#39;99\u0026#39;] Good Binary Strings: A binary string is a string consisting of 0\u0026rsquo;s and 1\u0026rsquo;s.\nA good binary string is defined as follows:\nThe number of 1\u0026rsquo;s is the same as the number of 0\u0026rsquo;s For every prefix in the string, the number of 1\u0026rsquo;s is not less than the number of 0\u0026rsquo;s You are given a good binary string. This string can contain multiple good binary substrings.\nIf two substrings are good and are adjacent (no overlap) to each other, they can be swapped. Perform zero or more swaps of such adjacent good substrings on the given binary string to get the lexicographically largest string.\nThey didn\u0026rsquo;t mention binary string here, so I\u0026rsquo;m assuming it is the lexicographically largest string.\nCoding Friends: Two friends Erica and Bob like to solve questions, each day they solve one question.\nThere are 3 types of questions:\nType Points E 1 M 3 H 5 Two lists are given that entail what problems both solve every day.\nPrint \u0026lsquo;Erica\u0026rsquo;, \u0026lsquo;Bob\u0026rsquo; or \u0026lsquo;Tied\u0026rsquo; depending on who has the higher points at the end of all the problems.\n","permalink":"https://manasch.github.io/placements/code/endor_labs/","summary":"Endor Labs Details Job Status Internship Only (With possible conversion based on performance)\nCriteria Study Cutoff X % XII % UG 8 GPA Round 1 04/11/23\nCoding Questions Log Files: A bank has transaction logs in the form of \u0026ldquo;sender_user_id receiver_user_id amount\u0026rdquo;. Given multiple such logs in an array and a threshold value of transactions, return a list of the user_id\u0026rsquo;s whose total transaction count is greater than the threshold in ascending order.","title":""},{"content":"Hiver Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 80% XII 80% UG 8 GPA Round 1 07/10/23\nThere were only 7 questions, 5 MCQ and 2 Coding, in 1h15m\nCoding Questions Next Smallest Palindrome: Given a number as an array, find the next smallest palindrome. vector\u0026lt;int\u0026gt; solve(vector\u0026lt;int\u0026gt; num) { bool isPalin = false; int n = num.size(); auto helper = [\u0026amp;] () { int l = 0; int r = n - 1; int t, oldr, p; while (l \u0026lt; r) { oldr = num[r]; num[r] = num[l]; if (oldr \u0026gt; num[l]) { t = r - 1; while (t \u0026gt; 0) { p = num[t]; ++p; if (p \u0026lt; 10) { num[t] = p; break; } else { num[t] = 0; } --t; } --l; ++r; } ++l; --r; } if (l \u0026gt;= r) { isPalin = true; } }; while (!isPalin) { helper(); } return num; } Covid Beds: There is a line of beds represented by 1\u0026rsquo;s and 0\u0026rsquo;s in a string, 1 means that the bed is occupied and 0 means that the bed is vacant. A partition is defined as such that it has exactly 2 occupied beds (this can also mean that the parition can have vacant beds too).\nA partition is created by placing a temporary walls between any two beds.\nPrint the number of ways the walls can be placed such that partitions satisfying the conditions can be made. If no such partitions can be made, print -1.\nEx:\n6 110011 Output: 3 - [11|0011, 110|011, 1100|11] 6 101101 Output: 1 - [101|101] 11 11001100011 Output: 12 ","permalink":"https://manasch.github.io/placements/code/hiver/","summary":"Hiver Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 80% XII 80% UG 8 GPA Round 1 07/10/23\nThere were only 7 questions, 5 MCQ and 2 Coding, in 1h15m\nCoding Questions Next Smallest Palindrome: Given a number as an array, find the next smallest palindrome. vector\u0026lt;int\u0026gt; solve(vector\u0026lt;int\u0026gt; num) { bool isPalin = false; int n = num.size(); auto helper = [\u0026amp;] () { int l = 0; int r = n - 1; int t, oldr, p; while (l \u0026lt; r) { oldr = num[r]; num[r] = num[l]; if (oldr \u0026gt; num[l]) { t = r - 1; while (t \u0026gt; 0) { p = num[t]; ++p; if (p \u0026lt; 10) { num[t] = p; break; } else { num[t] = 0; } --t; } --l; ++r; } ++l; --r; } if (l \u0026gt;= r) { isPalin = true; } }; while (!","title":""},{"content":"HPE Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 80% XII 80% UG 7 GPA Round 1 06/09/23\nThere were 4 sections:\nAptitude (15) - 20 min Technical (33) - 40 min Coding(2) - 30 min (language 1) Coding(2) - 30 min (language 2) The options for languages was - cpp, c, python, java\nCoding Questions CPP Coding Nuggets Palindrome Number: Given a number, a palindrome can be construced by add the reverse of the number to itself: 12 + 21 = 33. But this is not the case for all numbers for ex: 57. Find the first palindrome that occurs when performing these operations for the given number. Edge case would be when the given number itself is a palindrome.\nint reverse(int num) { int n = num; int res = 0; while (n \u0026gt; 0) { res = res * 10 + n % 10; n = n / 10; } if (res == num) { return 0; } return res; } int palindrome(int N) { int num = N; int t = reverse(num); if (t == 0) { num += num; t = reverse(num); } while (t != 0) { num += t; t = reverse(num); } return num; } int main() { int N; cin \u0026gt;\u0026gt; n; cout \u0026lt;\u0026lt; palindrome(N) \u0026lt;\u0026lt; endl; return 0; } Number of cars: A car requires 4 wheels (w), 2 chairs (c) and 1 body (b). Given w, c, b, find out the number of cars that can be manufactured. int main() { int w, c, b; cin \u0026gt;\u0026gt; w \u0026gt;\u0026gt; c \u0026gt;\u0026gt; b; int count = 0; while (w \u0026gt;= 4 \u0026amp;\u0026amp; c \u0026gt;= 2 \u0026amp;\u0026amp; b \u0026gt;= 1) { ++count; w -= 4; --b; c -= 2; } cout \u0026lt;\u0026lt; count \u0026lt;\u0026lt; endl; return 0; } Python Coding Nuggets Absolute Volume Difference: Given length, breadth and height of two cubes, find the absolue volume difference between the two. int difference(l1, b1, h1, l2, b2, h2): return abs(l1 * b1 * h1 - l2 * b2 * h2) New Array: Given an array, modify it such that the value at the kth position is the sum of the values of it\u0026rsquo;s next two neighbors. int newArray(arr, n): lastArr = arr.copy() for i in range(n): arr[i] = lastArr[(i + 1) % n] + lastArr[(i + 2) % n] ","permalink":"https://manasch.github.io/placements/code/hpe/","summary":"HPE Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 80% XII 80% UG 7 GPA Round 1 06/09/23\nThere were 4 sections:\nAptitude (15) - 20 min Technical (33) - 40 min Coding(2) - 30 min (language 1) Coding(2) - 30 min (language 2) The options for languages was - cpp, c, python, java\nCoding Questions CPP Coding Nuggets Palindrome Number: Given a number, a palindrome can be construced by add the reverse of the number to itself: 12 + 21 = 33.","title":""},{"content":"Hyperface Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 8 GPA Round 1 - Coding Round 18/08/23\nThere was only 1 question and we were given 2 hours.\nCoding Questions Chef\u0026rsquo;s Dishes: A chef wants to cook dishes, he gets one new ingredient each day, and this ingredient can be one of 5 types. (Protein, Carb, Fibre, Fat, Seasoning), and each type has an expiry date. The ingredient would be as follows: Fat1, CarbsRice, ProtienChicken, Fibre1\u0026hellip;\nFat - 1 (If received on Day 1, can be used on Day 1 or Day 2, but expires after Day 2) Carbs - 3 . .\nA dish is prepared when it meets the following conditions:\nIt uses exactly M ingredients It uses exactly N categories (Fat, Fibre etc). The expected output is as follows:\nIf the chef can cook on a day with the ingredients he has, print out the ingredients used in the order of their arrival separated by \u0026ldquo;:\u0026rdquo;. ex: Fat1:Fibre2:Carbs1. If the chef cannot cook then print \u0026ldquo;#\u0026rdquo;. If the chef can cook on a day he will mandatorily make the dish. Round 2 - Design Round Design an in-memory sophisticated cache that handles the following\nEviction Strategy: LRU, LFU. Write Policy: Write-Through, Write-Behind. TTL: Delete data from cache after ttl has expired. Cache Misses: Determine the number of cache misses at any point in time. Configurable cache size A two page document was provided with function declarations and details.\nCould pick any language of our choice, usage of internet was allowed but no LLM\u0026rsquo;s\n","permalink":"https://manasch.github.io/placements/code/hyperface/","summary":"Hyperface Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 8 GPA Round 1 - Coding Round 18/08/23\nThere was only 1 question and we were given 2 hours.\nCoding Questions Chef\u0026rsquo;s Dishes: A chef wants to cook dishes, he gets one new ingredient each day, and this ingredient can be one of 5 types. (Protein, Carb, Fibre, Fat, Seasoning), and each type has an expiry date.","title":""},{"content":"IBM Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 10/09/23\nThere was only one section with 2 coding questions. Everyone got 2 questions from a pool of questions.\nCoding Questions New Array: Given a list of integers and a list of intervals, for each interval, inverst the sign of the numbers with the interval, At the end return how the list would like like. (Follow 1-Based indexing).\nExample:\nL = [1, -4, 5, 3], intervals = [[2, 4], [1, 3]]\nFirst Interval : [1, 4, -5, 3]\nSecond Interval: [-1, -4, 5, -3]\nThe array after applying reversal of sign for all intervals should be returned.\nCircles Relationship: Given a list of strings such that each string contains the centres and radii of 2 circles, return a list of same length determining whether the circles: Intersect at two points, Touch, Concentric, Disjoint-Outside, Disjoint-Inside\nString would be: \u0026ldquo;x1 y1 r1 x2 y2 r1\u0026rdquo; and all the circle would either be centred on the x-axis or y-axis.\ndef euclidian(x1, y1, x2, y2): return ((x1 - x2) ** 2 + (y1 - y2) ** 2) ** 0.5 def solve(circles): for circle in circles: x1, y1, r1, x2, y2, r2 = [int(x) for x in circle.strip().split()] dist = euclidian(x1, y1, x2, y2) if x1 == x2 and y1 == y2: print(\u0026#34;Concentric\u0026#34;) if r1 + r2 == dist or max(r1, r2) - min(r1, r2) == dist: print(\u0026#34;Touching\u0026#34;) if max(r1, r2) - min(r1, r2) \u0026lt; d \u0026lt; r1 + r2: print(\u0026#34;Intersecting\u0026#34;) if r1 + r2 \u0026lt; dist: print(\u0026#34;Disjoint-Outside\u0026#34;) else: print(\u0026#34;Disjoint-Inside\u0026#34;) Compare JSON Given two json strings, find the difference between the common keys.\nNth Factor Given a number, return the Nth factor. Ex: K = 10, factors are: 1, 2, 5, 10, if n = 3, then 5 should be returned\nFind factors till sqrt(K) as the rest would repeat or be the inverse of the factors existing.\n","permalink":"https://manasch.github.io/placements/code/ibm/","summary":"IBM Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 10/09/23\nThere was only one section with 2 coding questions. Everyone got 2 questions from a pool of questions.\nCoding Questions New Array: Given a list of integers and a list of intervals, for each interval, inverst the sign of the numbers with the interval, At the end return how the list would like like.","title":""},{"content":"IDFC First Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 60% XII 60% UG 6 GPA Round 1 17/11/23\nThere were 3 sections.\nAptitude (15) - 20 min Technical (25) - 25 min (Had MS Excel and Stats too for some reason) Coding (2) - 45 min Coding Questions Distribute Cookies: There are N people sitting in a circle, the people are numbered from 1\u0026hellip;N. There\u0026rsquo;s a ball that player 1 holds and has to pass it on to other players. A random number K is selected which determines the distance any player has to pass the ball.\nWhen the ball returns to player 1, that round is over, and you have to print out the players who did not touch the ball.\nIf each player touched the ball before it goes back to 1, then print 0.\nvoid solve(int n, int k) { vector\u0026lt;bool\u0026gt; persons(n + 1, true); int person = 1; int count = 1; persons[person] = false; while ((person + k) % n != 1) { person = (person + k) % n; persons[person] = false; ++count; } if (count == n) { cout \u0026lt;\u0026lt; \u0026#34;0\u0026#34; \u0026lt;\u0026lt; endl; } else { for (int i = 1; i \u0026lt;= n; ++i) { if (persons[i]) { cout \u0026lt;\u0026lt; i \u0026lt;\u0026lt; \u0026#34; \u0026#34;; } } cout \u0026lt;\u0026lt; endl; } } Word Location: The input is a line of words seperated by space. You have to print the word and its position if the word starts with a vowel; if these words were sorted in ascending order.\nPrint 0 if there are no words that start with a vowel.\nbool isVowel(char ch) { return ch == \u0026#39;a\u0026#39; || ch == \u0026#39;e\u0026#39; || ch == \u0026#39;i\u0026#39; || ch == \u0026#39;o\u0026#39; || ch == \u0026#39;u\u0026#39; || ch == \u0026#39;A\u0026#39; || ch == \u0026#39;O\u0026#39; || ch == \u0026#39;I\u0026#39; || ch == \u0026#39;O\u0026#39; || ch == \u0026#39;U\u0026#39; } void solve(string words) { vector\u0026lt;string\u0026gt; wordss; string temp; for (char ch: words) { if (ch == \u0026#39; \u0026#39;) { wordss.push_back(temp); temp = \u0026#34;\u0026#34;; continue; } temp += ch; } wordss.push_back(temp); sort(wordss.begin(), wordss.end()); bool flag = true; for (int i = 0; i \u0026lt; n; ++i) { string word = wordss[i]; if (isVowel(word[0])) { flag = false; cout \u0026lt;\u0026lt; word \u0026lt;\u0026lt; \u0026#34; \u0026#34; \u0026lt;\u0026lt; i + 1 \u0026lt;\u0026lt; endl; } } if (flag) { cout \u0026lt;\u0026lt; \u0026#34;0\u0026#34; \u0026lt;\u0026lt; endl; } } Round 2 22/11/23\nThis was a technical round, was asked about OOPS, Strings, C++ (because it was in my resume)\n3 coding questions were asked\nGiven an array of numbers, and a value target, use the minimum number of numbers from the array to sum up to the target value. Find out if a number is an armstrong number or not. Find out if two strings are anagrams of each other. Round 3 22/11/23\nThis was a technical + HR round, it was different for different people.\nWas asked about projects in my resume and how oauth2 works, was given 1 coding question.\nGiven an array of characters which can contain duplicates, print the string which is the lexicographically largest and doesn\u0026rsquo;t contain any of the dupes. I told the set approach first, and then was asked to implement it without a set. Did it using sorting.\nOther basic HR questions.\nWhat do I know about IDFC? Why do I want to join IDFC? What have I done in the past 6 months to help in my career growth? How would you handle relocation and situations? ","permalink":"https://manasch.github.io/placements/code/idfc/","summary":"IDFC First Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 60% XII 60% UG 6 GPA Round 1 17/11/23\nThere were 3 sections.\nAptitude (15) - 20 min Technical (25) - 25 min (Had MS Excel and Stats too for some reason) Coding (2) - 45 min Coding Questions Distribute Cookies: There are N people sitting in a circle, the people are numbered from 1\u0026hellip;N. There\u0026rsquo;s a ball that player 1 holds and has to pass it on to other players.","title":""},{"content":"IDfy Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 6 GPA Round 1 16/10/23\nCoding Questions 3 Coding questions and 1 SQL\nCoding Preprocess Dates: Given a list of dates in the format 5-Mar-2023 convert it to 2023-03-05. unordered_map\u0026lt;string, string\u0026gt; months = { {\u0026#34;Jan\u0026#34;, \u0026#34;01\u0026#34;}, {\u0026#34;Feb\u0026#34;, \u0026#34;02\u0026#34;}, {\u0026#34;Mar\u0026#34;, \u0026#34;03\u0026#34;}, {\u0026#34;Apr\u0026#34;, \u0026#34;04\u0026#34;}, {\u0026#34;May\u0026#34;, \u0026#34;05\u0026#34;}, {\u0026#34;Jun\u0026#34;, \u0026#34;06\u0026#34;}, {\u0026#34;Jul\u0026#34;, \u0026#34;07\u0026#34;}, {\u0026#34;Aug\u0026#34;, \u0026#34;08\u0026#34;}, {\u0026#34;Sep\u0026#34;, \u0026#34;09\u0026#34;}, {\u0026#34;Oct\u0026#34;, \u0026#34;10\u0026#34;}, {\u0026#34;Nov\u0026#34;, \u0026#34;11\u0026#34;}, {\u0026#34;Dec\u0026#34;, \u0026#34;12\u0026#34;}, }; vector\u0026lt;string\u0026gt; preprocessDate(vector\u0026lt;string\u0026gt; dates) { vector\u0026lt;string\u0026gt; res; for (auto date: dates) { int i = 0; string day = \u0026#34;\u0026#34;; string month = \u0026#34;\u0026#34;; string year = \u0026#34;\u0026#34;; string t = \u0026#34;\u0026#34;; while (date[i] != \u0026#39; \u0026#39;) { if (date[i] - \u0026#39;0\u0026#39; \u0026lt; 10 \u0026amp;\u0026amp; date[i] - \u0026#39;0\u0026#39; \u0026gt;= 0) { t += date[i]; } ++i; } ++i; if (t.size() == 1) { day += \u0026#39;0\u0026#39;; } day += t; t = \u0026#34;\u0026#34;; while (date[i] != \u0026#39; \u0026#39;) { t += date[i]; ++i; } ++i; month = months[t]; t = \u0026#34;\u0026#34;; while (i \u0026lt; date.size()) { t += date[i]; ++i; } year = t; res.push_back(year + \u0026#34;-\u0026#34; + month + \u0026#34;-\u0026#34; + day); } return res; } Nuts \u0026amp; Bolts: There are two numbers, a secret number and a guess number, the number of digits in the guess number that are in the correct position of the secret number refer to as nuts.\nWhereas, the number of digits in the guess number that are in the secret number but at the wrong position is refered to as bots.\nPrint it as xAyB where x is nuts and y is bolts.\nstring solve(string secret, string guess) { int nuts = 0; int bolts = 0; int n = secret.size(); unordered_set\u0026lt;int\u0026gt; incorrect; unordered_map\u0026lt;char, int\u0026gt; remaining; for (int i = 0; i \u0026lt; n; ++i) { if (secret[i] == guess[i]) { ++nuts; } else { ++remaining[secret[i]]; incorrect.insert(i); } } for (int idx: incorrect) { if (remaining[guess[idx]] != 0) { ++bolts; --remaining[guess[idx]]; } } return to_string(nuts) + \u0026#34;A\u0026#34; + to_string(bolts) + \u0026#34;B\u0026#34;; } Error Logs: Given a list of error logs in the format of [date, time, type, status], return only those logs which are of type ERROR or CRITICAL and return them in the sorted order of date and time. If two logs appear at the same time, give priority to the one that appears first. string helper(string date, string time) { string year = \u0026#34;\u0026#34;; string month = \u0026#34;\u0026#34;; string day = \u0026#34;\u0026#34;; string t = \u0026#34;\u0026#34;; for (int i = 0; i \u0026lt; date.size(); ++i) { if (date[i] == \u0026#39;-\u0026#39;) { if (day == \u0026#34;\u0026#34;) { day = t; } else { month = t; } t.clear(); continue; } t += date[i]; } year = t; return year + month + day + time; } vector\u0026lt;vector\u0026lt;string\u0026gt;\u0026gt; extractErrorLogs(vector\u0026lt;vector\u0026lt;string\u0026gt;\u0026gt; logs) { vector\u0026lt;pair\u0026lt;string, int\u0026gt;\u0026gt; l; string log_string = \u0026#34;\u0026#34;; for (int i = 0; i \u0026lt; logs.size(); ++i) { auto log = logs[i]; if (log[2] == \u0026#34;ERROR\u0026#34; || log[2] == \u0026#34;CRITICAL\u0026#34;) { log_string += helper(log[0], log[1]); l.push_back({log_string, i}); } log_string.clear(); } sort(l.begin(), l.end()); vector\u0026lt;vector\u0026lt;string\u0026gt;\u0026gt; res; for (auto p: l) { res.push_back(logs[p.second]); } return res; } SQL CPU Usage: Two tables, servers (int id, varchar(255) cidr) and usage (int id, varchar(255) cpu_usage, varchar(255) network_usage, varchar(255) system_usage) are given.\nDisplay the server ip\u0026rsquo;s that have a cpu_usage of over a threshold, display the average of all the usages for all the listings for that server ip.\nThreshold: 80%\nServers\nid cidr 1 192.168.1.0/24 2 171.xxx.y.z/16 Usage\nid cpu_usage network_usage system_usage 1 1% 50% 7% 2 81% 5% x% 3 75% x% y% 2 23% x% y% 1 46% x% y% 2 95% x% y% 3 72% x% y% 2 23% x% y% id 2 has one or more listings with more than 80% cpu_usage so display the avg of all listings of id 2 Output\ncidr cpu_usage network_usage system_usage 171.xxx.y.z/16 66.34% x% y% ","permalink":"https://manasch.github.io/placements/code/idfy/","summary":"IDfy Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 6 GPA Round 1 16/10/23\nCoding Questions 3 Coding questions and 1 SQL\nCoding Preprocess Dates: Given a list of dates in the format 5-Mar-2023 convert it to 2023-03-05. unordered_map\u0026lt;string, string\u0026gt; months = { {\u0026#34;Jan\u0026#34;, \u0026#34;01\u0026#34;}, {\u0026#34;Feb\u0026#34;, \u0026#34;02\u0026#34;}, {\u0026#34;Mar\u0026#34;, \u0026#34;03\u0026#34;}, {\u0026#34;Apr\u0026#34;, \u0026#34;04\u0026#34;}, {\u0026#34;May\u0026#34;, \u0026#34;05\u0026#34;}, {\u0026#34;Jun\u0026#34;, \u0026#34;06\u0026#34;}, {\u0026#34;Jul\u0026#34;, \u0026#34;07\u0026#34;}, {\u0026#34;Aug\u0026#34;, \u0026#34;08\u0026#34;}, {\u0026#34;Sep\u0026#34;, \u0026#34;09\u0026#34;}, {\u0026#34;Oct\u0026#34;, \u0026#34;10\u0026#34;}, {\u0026#34;Nov\u0026#34;, \u0026#34;11\u0026#34;}, {\u0026#34;Dec\u0026#34;, \u0026#34;12\u0026#34;}, }; vector\u0026lt;string\u0026gt; preprocessDate(vector\u0026lt;string\u0026gt; dates) { vector\u0026lt;string\u0026gt; res; for (auto date: dates) { int i = 0; string day = \u0026#34;\u0026#34;; string month = \u0026#34;\u0026#34;; string year = \u0026#34;\u0026#34;; string t = \u0026#34;\u0026#34;; while (date[i] !","title":""},{"content":"Infibeam Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 80% XII 80% UG 8GPA There was another cut off, probably 8.5 GPA Round 1 07/10/23\nThere was only 1 section, 22 questions of which 2 were coding, a total of 1h45m was given to complete the test.\nCoding Questions Height of a tree: Given a binary tree, find the height of the tree by skipping the nodes that have odd-valued data. The height of an empty tree is 0 The height of a tree with just the root is also 0. If all the nodes are odd-valued, then the height is 0. int solve(Node *root) { int res = 0; auto dfs = [\u0026amp;] (auto self, Node *node, int height) { if (node == nullptr) { res = max(res, height); return; } if (node-\u0026gt;data % 2 == 0) { ++height; } self(self, node-\u0026gt;left, height); self(self, node-\u0026gt;right, height); }; dfs(dfs, root, 0); return max(res - 1, 0); } Word Search void solve(vector\u0026lt;vector\u0026lt;char\u0026gt;\u0026gt;\u0026amp; board, string word) { int m = board.size(); int n = board[0].size(); bool containsWord = false; visited\u0026lt;pair\u0026lt;int, int\u0026gt;\u0026gt; visited; auto bfs = [\u0026amp;] (auto self, int i, int j, int idx) { if (min(i, j) \u0026lt; 0 || i \u0026gt;= m || j \u0026gt;= n || containsWord) { return; } if (visited.find({i, j}) != visited.end()) { return; } if (idx == word.size()) { containsWord = true; return; } visited.insert({i, j}); if (board[i][j] == word[idx]) { self(self, i, j + 1); self(self, i, j - 1); self(self, i + 1, j); self(self, i - 1, j); } visited.erase({i, j}); }; for (int i = 0; i \u0026lt; m; ++i) { for (int j = 0; j \u0026lt; n; ++j) { bfs(bfs, i, j, 0); if (containsWord) { break; } } if (containsWord) { break; } } cout \u0026lt;\u0026lt; boolalpha \u0026lt;\u0026lt; res \u0026lt;\u0026lt; endl; } Round 2 11/10/23\nThis was a technical round, questions were asked based on resume.\nWhat is a stack? What is a linked list, hashtable? How is a hashtable implemented? What is an ordereddict? Time complexities. What is a heap, insertion time complexity? Difference between a heap and a binary tree? Project discussions. Code for insertion and finding an element in a linked list. Round 3 11/10/23\nThis was a coding/algo round\nWrite the code to flatten a binary tree into a linked list. Write the code to find the largest sum subarray. Leetcode ","permalink":"https://manasch.github.io/placements/code/infibeam/","summary":"Infibeam Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 80% XII 80% UG 8GPA There was another cut off, probably 8.5 GPA Round 1 07/10/23\nThere was only 1 section, 22 questions of which 2 were coding, a total of 1h45m was given to complete the test.\nCoding Questions Height of a tree: Given a binary tree, find the height of the tree by skipping the nodes that have odd-valued data.","title":""},{"content":"Intel Details There were two roles available:\nData Science: ML and Data Handling 2 open positions\nDescription: Work on automation of Data ETL for retail data\nDesired skills: Python, DL frameworks (PyTorch/TF), Data Engineering\nDeep Learning: Continuous Learning 1 open position\nDescription: CL algorithm basket and methodology addressing combination of non-IID data and \u0026gt;100K SKU’s\nDesired skills: Python, DL frameworks (PyTorch/TF), DL Training experience for Image\nJob Status 6-month Internship only (1 Jan 24 - 30 Jun 24)\nCriteria No criteria\nRound 1 19/09/23\nResume based shortlisting\nRound 2 (Final) 06/10/23\nThe interviewer focussed on asking questions based on previous experience and projects. As the role was AI/ML oriented, they were not asking coding questions.\n","permalink":"https://manasch.github.io/placements/code/intel/","summary":"Intel Details There were two roles available:\nData Science: ML and Data Handling 2 open positions\nDescription: Work on automation of Data ETL for retail data\nDesired skills: Python, DL frameworks (PyTorch/TF), Data Engineering\nDeep Learning: Continuous Learning 1 open position\nDescription: CL algorithm basket and methodology addressing combination of non-IID data and \u0026gt;100K SKU’s\nDesired skills: Python, DL frameworks (PyTorch/TF), DL Training experience for Image\nJob Status 6-month Internship only (1 Jan 24 - 30 Jun 24)","title":""},{"content":"Intuit Details Job status FTE + Internship (mandatory)\nCriteria Study Cutoff X % XII % UG 8 GPA Round 1 27/07/23\nQuestions Lost in Woods\nString parsing basically.\nInput\n12 S_**___*_**D \u0026lsquo;S\u0026rsquo; - start \u0026lsquo;D\u0026rsquo; - destination * - obstacle _ - path\nIf path exists can walk (W) else if an obstacle exists, can jump(J) over all the obstacles.\nOutput\nWJWWJJ Efficient Gardner Furious Dhoni\nN x M matrix and each cell contains a value ranging from -1000 to 1000, both N and M can range from 1 to 1000.\nGiven a ball which can start from any location, the ball can move to the right or down to its adjacent neighbor. If the ball goes from x1 y1 to x2 y2 (right or down), then the reward obtained is matrix[x2][y2] - matrix[x1][y1].\nFind the largest reward that can be obtained in the path where the ball moves only right or down.\nDon\u0026rsquo;t remember.. ","permalink":"https://manasch.github.io/placements/code/intuit/","summary":"Intuit Details Job status FTE + Internship (mandatory)\nCriteria Study Cutoff X % XII % UG 8 GPA Round 1 27/07/23\nQuestions Lost in Woods\nString parsing basically.\nInput\n12 S_**___*_**D \u0026lsquo;S\u0026rsquo; - start \u0026lsquo;D\u0026rsquo; - destination * - obstacle _ - path\nIf path exists can walk (W) else if an obstacle exists, can jump(J) over all the obstacles.\nOutput\nWJWWJJ Efficient Gardner Furious Dhoni\nN x M matrix and each cell contains a value ranging from -1000 to 1000, both N and M can range from 1 to 1000.","title":""},{"content":"Juniper Networks Details Job Status Internship + PPO\nCriteria Study Cutoff X % XII % UG 7.5 GPA Round 1 26/10/23\nThere were 14 questions in total to be solved in 110 min.\nMCQ (11) - Comprehension, Aptitude, Technical Coding (3) Coding Questions Decorate Function: Write a single decorator for 3 functions, which have variable arguements and one key value arguement delay.\nThere exists a global list execution_time which should include a number close to the delay_time for each query.\nThe function being decorated should return its respective value.\nExample:\n3 --\u0026gt; Number of queries delay_max 1 2 3 100 delay_min 3 4 1 7 200 delay_sum 2 2 300 Output 3 1 4 [100,200,300] # Pre-defined functions def delay_max(*args, **kwargs): time.sleep(kwargs[\u0026#34;delay\u0026#34;] / 1000) return max(args) def delay_min(*args, **kwargs): time.sleep(kwargs[\u0026#34;delay\u0026#34;] / 1000) return min(args) def delay_sum(*args, **kwargs): time.sleep(kwargs[\u0026#34;delay\u0026#34;] / 1000) return sum(args) execution_time = [] # # Complete the \u0026#39;timeit\u0026#39; function below. # # The function is expected to return a function. # The function accepts following parameter: # 1. FUNCTION func # def timeit(func: Callable) -\u0026gt; Callable: # Write your code here def wrapper(*args, **kwargs): execution_time.append(kwargs[\u0026#34;delay\u0026#34;]) return func(*args, **kwargs) return wrapper CPP Constructor: Given an empty class, this class will be called with the () operator, limit the number of times this operator is called.\nIf the number of calls exceeds this limit, throw -1 as the error.\nThe class also has a public method get_sum() which should return the total sum of all the arguements passed during the multiple () operator call.\nExample:\nxcallable obj_1(3); obj_1(3)(4)(5); // Output 12 xcallable obj_2(3); try { obj_2(3)(4)(5)(6); } catch (int x) { std::cout \u0026lt;\u0026lt; \u0026#34;Exception called\u0026#34; \u0026lt;\u0026lt; std::endl; } class xcallable{ private: int called = 0; int limit; int argsum = 0; public: xcallable(int limit) { this-\u0026gt;limit = limit; } xcallable\u0026amp; operator ()(int arg) { ++called; if (called \u0026gt; limit) { throw -1; } argsum += arg; return *this; } int get_sum() { return argsum; } }; Recover Dead Pods: There are n nodes from 1-n. Some of these nodes are connected to each other. This is represented with a list of undirected edges between two nodes. Nodes that are directly or indirectly connected to each other are part of the same region.\nEach node has their respective database connection, such that whenever a query is called, it will record it in their respective database.\nThere are q queries, each query of the following type:\n\u0026ldquo;1 pod_id\u0026rdquo;: Represents a normal query to the node pod_id. \u0026ldquo;2 pod_id\u0026rdquo;: Represents a disconnect to the database for the mentioned pod_id. (This disconnect is irreversible, once disconnected, cannot be reconnected) If a query is for a node whose database connection is down (due to a prior database disconnect query), this query is forwarded to the smallest valued node in the region the initial requested node belongs to and is written to that node\u0026rsquo;s database. Return the node which makes the database call. (Smallest if requested is down, else the requested node).\nIf the query is for a node whose database connection is down, and every other node in its region is also down, then the query is not written and is lost. In this case, return -1.\nExample:\n5 --\u0026gt; Nodes 3 --\u0026gt; Connections 1 2 2 3 4 5 6 --\u0026gt; Queries 1 4 2 2 1 2 2 3 2 1 1 1 Output: [4,1,-1] Explanation: 1 4 --\u0026gt; The requested node is alive, hence written to that database. 2 2 --\u0026gt; Node 2 lost its database connection. 1 2 --\u0026gt; Node 2 doesn\u0026#39;t have an active database connection, it belongs to the region whose smallest valued node is 1, hence written to 1. 2 3 --\u0026gt; Node 3 lost its database connection. 2 1 --\u0026gt; Node 1 lost its database connection. 1 1 --\u0026gt; Node 1 doesn\u0026#39;t have an active database connection and other nodes in its region are down, hence -1. class DSU { private: vector\u0026lt;int\u0026gt; parents, rank; int n; public: DSU(int n) { this-\u0026gt;n = n; parents = vector\u0026lt;int\u0026gt;(n + 1); rank = vector\u0026lt;int\u0026gt;(n + 1, 1); iota(parents.begin(), parents.end(), 0); } int find(int u) { if (parents[u] == u) { return u; } return parents[u] = find(parents[u]); } void join(int u, int v) { int up = find(u); int vp = find(v); if (up == vp) { return; } if (rank[up] \u0026gt; rank[vp]) { parents[vp] = up; } else { parents[up] = vp; ++rank[vp]; } } }; vector\u0026lt;int\u0026gt; recoverDeadPods(int n, vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt; connections, vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt; queries) { vector\u0026lt;int\u0026gt; res; DSU uf(n); for (auto\u0026amp; conn: connections) { uf.join(conn[0], conn[1]); } unordered_map\u0026lt;int, set\u0026lt;int\u0026gt;\u0026gt; regions; for (int i = 1; i \u0026lt;= n; ++i) { regions[uf.find(i)].insert(i); } vector\u0026lt;bool\u0026gt; alive(n + 1, true); // ---- for (auto\u0026amp; q: queries) { int qid = q[0]; int podId = q[1]; int parent = uf.find(podId); if (qid == 2) { alive[podId] = false; if (regions[parent].find(podId) != regions[parent].end()) { regions[parent].erase(podId); } } else { if (alive[podId]) { res.push_back(podId); } else { int smallestAlive = -1; if (!regions[parent].empty()) { smallestAlive = *regions[parent].begin(); } res.push_back(smallestAlive); } } } return res; } ","permalink":"https://manasch.github.io/placements/code/juniper_networks/","summary":"Juniper Networks Details Job Status Internship + PPO\nCriteria Study Cutoff X % XII % UG 7.5 GPA Round 1 26/10/23\nThere were 14 questions in total to be solved in 110 min.\nMCQ (11) - Comprehension, Aptitude, Technical Coding (3) Coding Questions Decorate Function: Write a single decorator for 3 functions, which have variable arguements and one key value arguement delay.\nThere exists a global list execution_time which should include a number close to the delay_time for each query.","title":""},{"content":"KeySight Details Job Status Internship Only\nWith possible conversion based on performance and vacancy Criteria Study Cutoff X 75% XII 75% UG 8 GPA Round 1 07/11/23\n40 MCQ\u0026rsquo;s on Aptitude, Networking and Programming in 60 min.\n","permalink":"https://manasch.github.io/placements/code/keysight/","summary":"KeySight Details Job Status Internship Only\nWith possible conversion based on performance and vacancy Criteria Study Cutoff X 75% XII 75% UG 8 GPA Round 1 07/11/23\n40 MCQ\u0026rsquo;s on Aptitude, Networking and Programming in 60 min.","title":""},{"content":"MakeMyTrip Details Job Status 6 Months Internship + PPO\nCriteria Study Cutoff X % XII % UG 7 GPA Round 1 16/09/23\nThere were 17 Questions to be solved in 1.5 hours\nMCQ (15) Coding (2) Coding Questions Telephone Lines: There are M houses and N localities, your job as a city planner is to design a plan such that atleast one house exists in a locality.\nIn a locality, all the houses have a telephone line connected to each other (n*(n-1)/2). Generate two plans to find out the maximum and minimum number of telephone lines that will be required across all the localities.\nMax would be generated when all localities except one has 1 house, and the rest in the last locality. The min would be when the houses are equally distributed across all localities.\nNew Neighbours: There are M houses built in a row and there are N people wanting to move into the houses. (N \u0026lt;= M). Everyone wants to stay as far away as possible from each other, what would be the maximised minimum distance between the houses.\nAn array of size M is provided which has the distances from the front gate. Assign the N people to the houses such that they are as far away as possible from each other.\nPerform a binary search to determine the distance between the houses and keep updating the minimum distance.\n","permalink":"https://manasch.github.io/placements/code/makemytrip/","summary":"MakeMyTrip Details Job Status 6 Months Internship + PPO\nCriteria Study Cutoff X % XII % UG 7 GPA Round 1 16/09/23\nThere were 17 Questions to be solved in 1.5 hours\nMCQ (15) Coding (2) Coding Questions Telephone Lines: There are M houses and N localities, your job as a city planner is to design a plan such that atleast one house exists in a locality.\nIn a locality, all the houses have a telephone line connected to each other (n*(n-1)/2).","title":""},{"content":"Microchip Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 17/10/23\n60 MCQ\u0026rsquo;s based on aptitude, technical and electrical knowledge in 70 min.\n","permalink":"https://manasch.github.io/placements/code/microchip/","summary":"Microchip Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 17/10/23\n60 MCQ\u0026rsquo;s based on aptitude, technical and electrical knowledge in 70 min.","title":""},{"content":"Morgan Stanley Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 8 GPA Round 1 17/09/23\nThere were 3 sections and a total of 100 min.\nAutomata Fix (7) - 20min - (A debugging section, debug code such that it passes all test cases.) Quantitative and Logical Aptitude (10) - 20min Coding (3) - 60min Coding Questions Longest Common Prefix: Given an array of names, find the longest common prefix of all the names. string solve(vector\u0026lt;string\u0026gt;\u0026amp; names, int n) { if (n == 0) { return \u0026#34;\u0026#34;; } if (n == 1) { return names[0]; } sort(names.begin(), names.end()); string min_string = names.front().size() \u0026lt; names.back().size() ? names.front() : names.back(); int min_length = min_string.size(); int i = 0; string res = \u0026#34;\u0026#34;; while (i \u0026lt; min_length \u0026amp;\u0026amp; names.front()[i] == names.back()[i]) { res += min_string[i]; ++i; } return res; } int main() { int n; cin \u0026gt;\u0026gt; n; vector\u0026lt;string\u0026gt; names(n); for (int i = 0; i \u0026lt; n; ++i) { cin \u0026gt;\u0026gt; names[i]; transform(names[i].begin(), names[i].end(), names[i].begin(), [](unsigned char c) {return tolower(c);}); } string res = solve(names, n); cout \u0026lt;\u0026lt; res \u0026lt;\u0026lt; endl; return 0; } GFG Max Substring Length: Given a number as string, find the largest length of a substring of size 2k such that the sum of left k digits is equal to the sum of the right k digits. num = input() n = len(num) maxLen = 0 for i in range(n - 1): l = i r = i + 1 leftSum = 0 rightSum = 0 while (l \u0026gt;=0 and r \u0026lt; n): leftSum += int(num[l]) rightSum += int(num[r]) if leftSum == rightSum: maxLen = max(maxLen, r - l + 1) l -= 1 r += 1 print(maxLen) int solve(string num, int n) { int l, r, leftSum, rightSum, maxLen = 0; for (int i = 0; i \u0026lt; n - 1; ++i) { l = i; r = i + 1; leftSum = 0; rightSum = 0; while (l \u0026gt;= 0 \u0026amp;\u0026amp; r \u0026lt; n) { leftSum += num[l] - \u0026#39;0\u0026#39;; rightSum += num[r] - \u0026#39;0\u0026#39;; if (leftSum == rightSum) { maxLen = max(maxLen, r - l + 1); } --l; ++r; } } return maxLen; } int main() { string num; cin \u0026gt;\u0026gt; num; int res = solve(num, num.size()); cout \u0026lt;\u0026lt; res \u0026lt;\u0026lt; endl; return 0; } Number of Different Plants: Given a 2D matrix of 0\u0026rsquo;s and 1\u0026rsquo;s, find the different number plants that exist. In the cell, if a plant exists, it is marked as 1 otherwise 0.\nThe plants that are connected 4-sided are of the same type, whereas the plants that are connected diagonally are considered different plants. Hence given these conditions, find the total number of different plants.\nint solve(vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt;\u0026amp; matrix, int m, int n) { int count = 0; set\u0026lt;pair\u0026lt;int, int\u0026gt;\u0026gt; visited; auto bfs = [\u0026amp;] (auto self, int i, int j) { if (min(i, j) \u0026lt; 0 || i \u0026gt;= m || j \u0026gt;= n) { return; } if (matrix[i][j] == 0 || visited.find({i, j}) != visited.end()) { return; } visited.insert({i, j}); self(self, i, j + 1); self(self, i + 1, j); self(self, i, j - 1); self(self, i - 1, j); }; for (int i = 0; i \u0026lt; m; ++i) { for (int j = 0; j \u0026lt; n; ++j) { if (matrix[i][j] == 1 \u0026amp;\u0026amp; visited.find({i, j}) == visited.end()) { bfs(bfs, i, j); ++count; } } } return count; } int main() { int n, m; cin \u0026gt;\u0026gt; n \u0026gt;\u0026gt; m; vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt; graph(n, vector\u0026lt;int\u0026gt;(m, 0)); for (int i = 0; i \u0026lt; n; ++i) { for (int j = 0; j \u0026lt; m; ++j) { cin \u0026gt;\u0026gt; graph[i][j]; } } int ans = solve(graph, n, m); cout \u0026lt;\u0026lt; ans \u0026lt;\u0026lt; endl; return 0; } This is exactly LC#200 but worded differently. ","permalink":"https://manasch.github.io/placements/code/morgan_stanley/","summary":"Morgan Stanley Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 8 GPA Round 1 17/09/23\nThere were 3 sections and a total of 100 min.\nAutomata Fix (7) - 20min - (A debugging section, debug code such that it passes all test cases.) Quantitative and Logical Aptitude (10) - 20min Coding (3) - 60min Coding Questions Longest Common Prefix: Given an array of names, find the longest common prefix of all the names.","title":""},{"content":"Natwest Group Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 6 GPA Round 1 - Coding Round 29/08/23\nThere were multiple sections, technical, aptitude, english, analytical. 2 Coding Everyone had different coding questions.\nCoding Questions Characters in same position: Given a string, find out the number of characters that will be in the same position once the string is reversed. Unique Elements: Given a list of integers, return the unique elements in the order they appear. Round 2 - Interview There was only 1 round for everyone. About yourself, hobbies and tech stack used. Projects explanation from resume or other interests. Pseudocode on any one sorting algo. Why Natwest?? ","permalink":"https://manasch.github.io/placements/code/natwest/","summary":"Natwest Group Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 6 GPA Round 1 - Coding Round 29/08/23\nThere were multiple sections, technical, aptitude, english, analytical. 2 Coding Everyone had different coding questions.\nCoding Questions Characters in same position: Given a string, find out the number of characters that will be in the same position once the string is reversed. Unique Elements: Given a list of integers, return the unique elements in the order they appear.","title":""},{"content":"NVIDIA Details Job Status 6 Month Internship\nCriteria Study Cutoff X % XII % UG 8 GPA Round 1 20/09/23\n3 Sections, 60 min and not timed by section.\nAnalytical (15) Technical (12) Coding (2) Coding Questions Last Stone Weight int lastStoneWeight(vector\u0026lt;int\u0026gt;\u0026amp; stones) { priority_queue\u0026lt;int, vector\u0026lt;int\u0026gt;\u0026gt; q(stones.begin(), stones.end()); int s, t; while (q.size() \u0026gt; 1) { s = q.top(); q.pop(); t = q.top(); q.pop(); if ((s - t) != 0) { q.push(s - t); } } if (q.size() \u0026gt; 0) { return q.top(); } return 0; } Regex Match: Given a binary representation of a number in string, give the regex pattern to determine if it is a power of 2. This was weird, CPP12 language selection had the parsing code already and we just had to mention the regex pattern, whereas CPP14 and CPP20 language selection, we had to take the input and perform the parsing. Nonetheless, parsing can be done with the help of a flag.\nPower of 2 will always have one 1 and 0\u0026#39;s on both sides. 0*10* ","permalink":"https://manasch.github.io/placements/code/nvidia/","summary":"NVIDIA Details Job Status 6 Month Internship\nCriteria Study Cutoff X % XII % UG 8 GPA Round 1 20/09/23\n3 Sections, 60 min and not timed by section.\nAnalytical (15) Technical (12) Coding (2) Coding Questions Last Stone Weight int lastStoneWeight(vector\u0026lt;int\u0026gt;\u0026amp; stones) { priority_queue\u0026lt;int, vector\u0026lt;int\u0026gt;\u0026gt; q(stones.begin(), stones.end()); int s, t; while (q.size() \u0026gt; 1) { s = q.top(); q.pop(); t = q.top(); q.pop(); if ((s - t) != 0) { q.","title":""},{"content":"Pure Storage Details Job Status FTE + Internship (Mandatory)\nCriteria Study Cutoff X % XII % UG 7GPA Round 1 02/08/23\nThe questions were different for everyone.\nTwo sections\nMCQ (11) Coding (2) Coding Questions Minimum Operations to Make the Array Alternating This but the input array was just of 0\u0026rsquo;s and 1\u0026rsquo;s\nPython Requests Library Had to use python\u0026rsquo;s request module to fetch data from an api endpoint and give a summary\nLongest Increasing Subsequence ","permalink":"https://manasch.github.io/placements/code/pure_storage/","summary":"Pure Storage Details Job Status FTE + Internship (Mandatory)\nCriteria Study Cutoff X % XII % UG 7GPA Round 1 02/08/23\nThe questions were different for everyone.\nTwo sections\nMCQ (11) Coding (2) Coding Questions Minimum Operations to Make the Array Alternating This but the input array was just of 0\u0026rsquo;s and 1\u0026rsquo;s\nPython Requests Library Had to use python\u0026rsquo;s request module to fetch data from an api endpoint and give a summary","title":""},{"content":"PwC Acceleration Centers Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 60%/6 GPA Cyber Security as Major or Elective Round 1: Cyber Recruit Boot Camp 08/08/23\nThere were 4 sections:\nComputer Fundamental MCQs (16) Cybersecurity MCQs (16) Coding Question - Easy (1) Coding Question - Novice (1) Coding Questions had a choice of C, Java and Python. Questions were different for everyone.\nQuestions Coding Question - Easy Remove duplicates from string A function to remove all duplicate alphabets and non-alphabetic characters from a string and sort it in ascending order. The resultant string was supposed to be lower-case.\ndef solve(str): str = sorted(str.lower()) res = \u0026#34;\u0026#34; for i in str: if i.isalpha() and i not in res: res += i return res Coding Question - Novice Special Number A function to return \u0026ldquo;Special number\u0026rdquo; if any of the digits in the number were perfect squares of 1, 2 or 3, otherwise return \u0026ldquo;Not a special number\u0026rdquo;\ndef solve(n): while n != 0: if n % 10 in [1,4,9]: return \u0026#34;Special number\u0026#34; n //= 10 return \u0026#34;Not a special number\u0026#34; ","permalink":"https://manasch.github.io/placements/code/pwc/","summary":"PwC Acceleration Centers Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 60%/6 GPA Cyber Security as Major or Elective Round 1: Cyber Recruit Boot Camp 08/08/23\nThere were 4 sections:\nComputer Fundamental MCQs (16) Cybersecurity MCQs (16) Coding Question - Easy (1) Coding Question - Novice (1) Coding Questions had a choice of C, Java and Python. Questions were different for everyone.","title":""},{"content":"Ring Central Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 90% XII 90% UG 7.5 Round 1 06/08/23\nTwo sections\nMCQ (15) Coding (2) MCQ Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Coding Q16 House Robber on Leetcode\nQ17 ","permalink":"https://manasch.github.io/placements/code/ring_central/","summary":"Ring Central Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 90% XII 90% UG 7.5 Round 1 06/08/23\nTwo sections\nMCQ (15) Coding (2) MCQ Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Coding Q16 House Robber on Leetcode\nQ17 ","title":""},{"content":"Sahaj Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 24/08/23\nOnline test based on fundamentals - CN, OS, DBMS, OOPS. A total of 20 questions.\nRound 2 - Building in-memory DB for backend server. Problem statement We need a system to be able to manage data for employees. Requests will be accepted over HTTP (API contract). No databases/libraries can be used to store/maintain data.\nFastest applications win.\nTechnical details Your repository needs to have a Dockerfile that starts your HTTP web app Your HTTP app need to expose APIs (API contract) on port 8080 No existing databases, libraries and services can be used to store the data Application needs to persist data across restarts No limitation on the programming language Do not touch the GitHub actions code. It is used to test your code automatically and score it. Any modifications will lead to immediate disqualification. Maximum time a single request can take is 10 seconds Data should be persisted in /home/ FAQ Do not run a development webserver with watch enabled in your app. Your tests will fail. If your greeting end point test is not passing, please check the output you produce. It needs to be exactly what is requested. When in doubt, please check the Github Actions logs for details Logs for the performance tests will not be shared Data to be stored { employeeId: string, name: string, city: string } API contract GET /greeting Checks whether the service is available.\nResponse Code: 200 Content: Hello world! POST /employee Creates a new Employee and returns the employeeId\nRequest \u0026amp; Response headers Content-Type: application/json\nBody { name: string, city: string } Success Response Status code: 201 Content: { \u0026quot;employeeId\u0026quot;: \u0026quot;\u0026lt;employee_id\u0026gt;\u0026quot; } (Note: Employee ID is a string) GET /employee/:id Returns the specified employee.\nURL Params id=[string] Required\nSuccess Response Status code: 200 Content: { \u0026lt;employee_object\u0026gt; } Error Response Status code: 404 Content: { message : \u0026quot;Employee with \u0026lt;employee_id\u0026gt; was not found\u0026quot; } GET /employees/all Returns list of all employees.\nSuccess Response Status code: 200 Content: [{ \u0026lt;employee_object\u0026gt; }] PUT /employee/:id Updates fields of the existing employee and returns the new object.\nURL Params id=[string] Required\nHeaders Content-Type: application/json\nBody { name: string, city: string } Success Response Code: 201 Content: { \u0026lt;employee_object\u0026gt; } Error Response Code: 404 Content: { message : \u0026quot;Employee with \u0026lt;employee_id\u0026gt; was not found\u0026quot; } DELETE /employee/:id Deletes existing employee record.\nURL Params id=[string] Required\nSuccess Response Status code: 200 Content: { \u0026lt;employee_object\u0026gt; } Error Response Status code: 404 Content: { message : \u0026quot;Employee with \u0026lt;employee_id\u0026gt; was not found\u0026quot; } Competition rules Check rules and scoring pages for details. When in doubt, ask the organizers and we will add clarifications to the page.\nSample Instructions for coding/committing git clone git repository url (Skip this step if using github codespaces) cd repository name (Skip this step if using github codespaces) Goto server.js Begin coding Code needs to be tested via github? Execute following commands in the terminal location of your repository git add . git commit -m \u0026lsquo;any message regarding your changes\u0026rsquo; git push Wait for build to complete in github actions logs. If build is green, you should see the score on the leader board, else check with actions logs. Sample Instructions for Codespaces installation On your browser after accepting the github invitation, Celect \u0026ldquo;Code\u0026rdquo; dropdown Select the \u0026ldquo;Codespaces\u0026rdquo; tab. Select \u0026ldquo;Create codespace on main\u0026rdquo; Continue from step 3 of Sample Instructions for coding/commit ","permalink":"https://manasch.github.io/placements/code/sahaj/","summary":"Sahaj Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 24/08/23\nOnline test based on fundamentals - CN, OS, DBMS, OOPS. A total of 20 questions.\nRound 2 - Building in-memory DB for backend server. Problem statement We need a system to be able to manage data for employees. Requests will be accepted over HTTP (API contract). No databases/libraries can be used to store/maintain data.","title":""},{"content":"Samsung Research Institute Details Job Status Internship Only\nCriteria Study Cutoff X % XII % UG 7 GPA Round 1 18/10/23\nThere were 3 coding questions to be solved in 80 min, everyone got 3 questions from a pool of questions.\nCoding Questions Christmas Gifts: There are N chocolates and M toys, a gift box can have either X chocolates and Y toys or X toys and Y chocolates.\nFind the maximum number of gift boxes that can be made.\nint solve(int n, int m, int x, int y) { vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt; dp(n + 1, vector\u0026lt;int\u0026gt;(m + 1, -1)); auto dfs = [\u0026amp;] (auto self, int i, int j) { if (min(i, j) \u0026lt; 0) { return -1; } if (dp[i][j] != -1) { return dp[i][j]; } int left = self(self, i - x, j - y); int right = self(self, i - y, j - x); return dp[i][j] = 1 + max(right, left); }; return dfs(dfs, n, m); } Hamiltonian Circuits: Given an undirected graph, return the total number of unique hamiltonian circuits.\nNote: An hamiltonian circuit is a path in which each node in the graph is visited once except for the source node which is visited twice.\nGFG Worms: Given some number of worms and some number of interactions amongst the worms, figure out whether these interactions are valid or not.\nA worm can be a male or female. A valid interaction is when a male worm interacts with the female worm and vice versa.\nFor the given set of interactions, figure out if it is a valid set of interactions or an invalid set of interactions.\nEx:\n3 (worms) 3 (interactions) 1 2 2 3 1 3 This is invalid as 1 interacts with 2 and 3 but 2 interacts with 3, there is no possible combination of genders where this is valid. ","permalink":"https://manasch.github.io/placements/code/samsung_rnd/","summary":"Samsung Research Institute Details Job Status Internship Only\nCriteria Study Cutoff X % XII % UG 7 GPA Round 1 18/10/23\nThere were 3 coding questions to be solved in 80 min, everyone got 3 questions from a pool of questions.\nCoding Questions Christmas Gifts: There are N chocolates and M toys, a gift box can have either X chocolates and Y toys or X toys and Y chocolates.\nFind the maximum number of gift boxes that can be made.","title":""},{"content":"Samsung Semiconductor India Research Details Job Status Internship Only\nCriteria Study Cutoff X % XII % UG 7 GPA No Backlogs Round 1 05/10/23\nOne section with 3 coding questions for 70 min.\nCoding Questions Alternate Levels: Given the root of a binary tree, return the sum of all the nodes in the alternate levels of a binary tree starting from the root. If the root is null then return 0. int solve(TreeNode *root) { if (root == nullptr) { return 0; } int res = 0; int level = 0; queue\u0026lt;TreeNode *\u0026gt; q; q.push(root); while (!q.empty()) { int n = q.size(); while (n--) { TreeNode *t = q.front(); q.pop(); if (level % 0 == 0) { res += t-\u0026gt;data; } if (t-\u0026gt;left) { q.push(t-\u0026gt;left); } if (t-\u0026gt;right) { q.push(t-\u0026gt;right); } } ++level; } return res; } Number of Components: Given an m*n binary matrix, a connected component is a set of 1\u0026rsquo;s that can be reached 8-directionally. Return the number of connected components. This is basically Number of Islands on LeetCode but 8-dxns\nint solve(int **graph, int m, int n) { set\u0026lt;pair\u0026lt;int, int\u0026gt;\u0026gt; visited; auto dfs = [\u0026amp;] (auto self, int i, int j) { if (min(i, j) \u0026lt; 0 || i \u0026gt;= m || j \u0026gt;= n) { return; } if (visited.find({i, j}) != visited.end() || graph[i][j] == 0) { return; } visited.insert({i, j}); self(self, i, j + 1); self(self, i + 1, j); self(self, i, j - 1); self(self, i - 1, j); self(self, i - 1, j + 1); self(self, i - 1, j - 1); self(self, i + 1, j + 1); self(self, i + 1, j - 1); } for (int i = 0; i \u0026lt; m; ++i) { for (int j = 0; j \u0026lt; n; ++j) { if (visited.find({i, j}) == visited.end() \u0026amp;\u0026amp; graph[i][j] != 0) { dfs(dfs, i, j); ++res; } } } return res; } Next Palindrome: Given a number as a string, find the first number greater than the current one that is a palindrome. void helper(char *num, int n, bool *isPalin) { int l = 0; int r = n - 1; int p, t, oldr; char c; while (l \u0026lt; r) { if (num[l] != num[r]) { oldr = num[r]; num[r] = num[l]; if (oldr \u0026gt; num[l]) { t = r - 1; while (t \u0026gt; 0) { c = num[t]; p = atoi(c); ++p; if (p \u0026lt; 10) { num[t] = p + \u0026#39;0\u0026#39;; break; } else { num[t] = \u0026#39;0\u0026#39;; } --t; } } } ++l; --r; } if (l \u0026gt;= r) { isPalin = true; } } int solve(char *num) { int len = 0; while (num[len] != \u0026#39;\\0\u0026#39;) { ++len; } bool isPalin = false; while (!isPalin) { helper(num, len, \u0026amp;isPalin); } return num; } Not sure if this would pass hidden cases, did pass the visible ones.\n","permalink":"https://manasch.github.io/placements/code/samsung_semiconductor/","summary":"Samsung Semiconductor India Research Details Job Status Internship Only\nCriteria Study Cutoff X % XII % UG 7 GPA No Backlogs Round 1 05/10/23\nOne section with 3 coding questions for 70 min.\nCoding Questions Alternate Levels: Given the root of a binary tree, return the sum of all the nodes in the alternate levels of a binary tree starting from the root. If the root is null then return 0.","title":""},{"content":"Sandvine Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 7.5 GPA Full Time: Training Cost Recovery of Rs.2,00,000 if candidate leaves in one year of time\nInternship: Training Cost Recovery of Rs. 1,00,000 if candidate leaves in 6 months’ time\nRound 1 - Coding Round 22/08/23\n3 Sections;\nAptitude (15) MCQ (27?) Algorithm (3) Coding but no test cases, text box to type the algo or code Coding Questions Elevator: An elevator can go to any of 5 floors up at once, if it goes 5 floors up, then it has to go 3 down, find the minimum number of steps it requires to reach any particular floor. BT to BST: Write code to convert a Binary Tree to Binary Search Tree Hashing Servers: Write a template to distribute user request network traffic across multiple servers. Round 2 - Technical Interview About Me Technical Databases Compression Algos Big Table vs RDBMS Discuss code written during round 1. Round 3 - HR / Technical Interview HR Questions Resume Based. ","permalink":"https://manasch.github.io/placements/code/sandvine/","summary":"Sandvine Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 7.5 GPA Full Time: Training Cost Recovery of Rs.2,00,000 if candidate leaves in one year of time\nInternship: Training Cost Recovery of Rs. 1,00,000 if candidate leaves in 6 months’ time\nRound 1 - Coding Round 22/08/23\n3 Sections;\nAptitude (15) MCQ (27?) Algorithm (3) Coding but no test cases, text box to type the algo or code Coding Questions Elevator: An elevator can go to any of 5 floors up at once, if it goes 5 floors up, then it has to go 3 down, find the minimum number of steps it requires to reach any particular floor.","title":""},{"content":"Schneider Electric Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 28/08/23\nOnline Assessment MCQ\u0026rsquo;s based on aptitude, analytical, electronics, technical. Round 2 Technical Interview DSA question: How to detect a cycle in a linked list Rate yourself on a scale of 1 - 10 for a given techstack (MERN/MEAN), questions asked accordingly. Database design question Round 3 HR Interview HR Questions Resume Based. ","permalink":"https://manasch.github.io/placements/code/schneider_electric/","summary":"Schneider Electric Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG GPA Round 1 28/08/23\nOnline Assessment MCQ\u0026rsquo;s based on aptitude, analytical, electronics, technical. Round 2 Technical Interview DSA question: How to detect a cycle in a linked list Rate yourself on a scale of 1 - 10 for a given techstack (MERN/MEAN), questions asked accordingly. Database design question Round 3 HR Interview HR Questions Resume Based.","title":""},{"content":"Societe Generalte Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 60% XII 60% UG 6 GPA Round 1 06/09/23\nCoding Questions Number of Pairs: Given a string of characters, find the total number of pairs that can be formed of the same character. Ex: aaaaabb can have 2 pairs of a and 1 pair of b. Min Swap Greatest Even Number: Given a number as a string, find the greatest even number that can be made with just one swap, it is given that atleast one even digit exists. If the number is odd, then the last digit has to be switched with some other, so find the first even number the odd number is greater than if not, swap it with the rightmost even number, if it is even, then find two numbers two swap such that it is the greatest.\n","permalink":"https://manasch.github.io/placements/code/societe_generale/","summary":"Societe Generalte Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 60% XII 60% UG 6 GPA Round 1 06/09/23\nCoding Questions Number of Pairs: Given a string of characters, find the total number of pairs that can be formed of the same character. Ex: aaaaabb can have 2 pairs of a and 1 pair of b. Min Swap Greatest Even Number: Given a number as a string, find the greatest even number that can be made with just one swap, it is given that atleast one even digit exists.","title":""},{"content":"Sprinklr Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 6 GPA Round 1 02/09/23\nMCQ (72) - Technical, Aptitude Coding (2) Coding Questions Smallest Common Substring: Given two strings, find the number of mappings of the common substrings between the two stirngs. Don\u0026rsquo;t have to consider subtrings, since they\u0026rsquo;re asking for smallest, it just becomes the individual characters. Create a charCount for both and multiply and add the count for each character.\nSubstrings ending with same characters: Given a string, and p ranges (S[1..2] or S[1..6]). Find the number of substring in the ranges such that the end characters are the same. ","permalink":"https://manasch.github.io/placements/code/sprinklr/","summary":"Sprinklr Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 6 GPA Round 1 02/09/23\nMCQ (72) - Technical, Aptitude Coding (2) Coding Questions Smallest Common Substring: Given two strings, find the number of mappings of the common substrings between the two stirngs. Don\u0026rsquo;t have to consider subtrings, since they\u0026rsquo;re asking for smallest, it just becomes the individual characters. Create a charCount for both and multiply and add the count for each character.","title":""},{"content":"Swym Details Job Status (Internship + Full Time Employment Mandatory)\nCriteria None mentiond\nRound 1 09/08/23\nFirst round was conducted in 4 batches for Associate Software Engineer, Customer Experience Engineer was on 10/08/23. Each batch had different questions, some did repeat though.\nThere were 4 sections.\nScenario Based Questions (6) Aptitude MCQ Coding (3) Coding Questions Captcha Validity: Given N string, the captcha is valid if for each string, the frequence of each unique character in the string is the same. Otherwise invalid. Profile Picture: Given S, W, H, if either W or H is less than S, print \u0026ldquo;CHANGE\u0026rdquo;, otherwise if it\u0026rsquo;s a square print \u0026ldquo;ACCEPTED\u0026rdquo; else print \u0026ldquo;CROP\u0026rdquo;. 2D Array Less Than: Given a 2D matrix, return count of for i \u0026lt; j, arr[i] \u0026gt; arr[j]. Infinite 1D Jungle: Given M, choose starting position from 1 - M (included) such that most number of places is visited. For all N, if N is even, go to N / 2, if N is odd go to 3 * N + 1. Round 2 (ASE) 10/08/23\nThe second round for ASE was in the form of an assignment. The shortlisted candidates were told to choose one of two assignment statements, and given 4 hours to complete the code, create a repo and 1 additional hour to create a video demo and a README file containing:\nexplanation technologies used how to install and run At the end of this 5-hour round, best solutions were selected and the candidate was asked to present to a panel of judges. This was followed by HR round and then results.\nAssignments The two assignment choices given were different for each candidate, but the themes were the same:\nAssignment 1: Database + UI Assignment 2: Data Processing + ML/Data Analytics ","permalink":"https://manasch.github.io/placements/code/swym/","summary":"Swym Details Job Status (Internship + Full Time Employment Mandatory)\nCriteria None mentiond\nRound 1 09/08/23\nFirst round was conducted in 4 batches for Associate Software Engineer, Customer Experience Engineer was on 10/08/23. Each batch had different questions, some did repeat though.\nThere were 4 sections.\nScenario Based Questions (6) Aptitude MCQ Coding (3) Coding Questions Captcha Validity: Given N string, the captcha is valid if for each string, the frequence of each unique character in the string is the same.","title":""},{"content":"Tesco Details Job Status Full Time (Employment only)\nCriteria Study Cutoff X % XII % UG 8 GPA Round 1 10/08/23\nHad 3 sections\nAptitude MCQ (25) - 30 min MCQ (10) - 20 min Coding (2) - 40 min Coding Questions Don\u0026rsquo;t print rows containing -1: Given a square 2D array of integers, don\u0026rsquo;t print rows that contain -1 Index at which it becomes greater: Given N and P, form a circle of all the primes under N (including) and start from 2 and keep adding values in clockwise manner till the sum becomes greater than P, return the index at which the sum crosses P. void generatePrimes(int n, vector\u0026lt;int\u0026gt;\u0026amp; primes) { primes.clear(); vector\u0026lt;bool\u0026gt; isPrime(n + 1, true); for (int i = 2; i * i \u0026lt;= n; ++i) { if (isPrime[i]) { for (int j = i * i; j \u0026lt;= n; j += i) { isPrime[j] = false; } } } for (int i = 2; i \u0026lt;= n; ++i) { if (isPrime[i]) { primes.push_back(i); } } } int solve(int n, int p) { vector\u0026lt;int\u0026gt; primes; generatePrimes(n, primes); int idx = 0; int s = 0; while (s \u0026gt; p) { s += primes[idx % primes.size()]; ++idx; } return (idx - 1) % primes.size(); } ","permalink":"https://manasch.github.io/placements/code/tesco/","summary":"Tesco Details Job Status Full Time (Employment only)\nCriteria Study Cutoff X % XII % UG 8 GPA Round 1 10/08/23\nHad 3 sections\nAptitude MCQ (25) - 30 min MCQ (10) - 20 min Coding (2) - 40 min Coding Questions Don\u0026rsquo;t print rows containing -1: Given a square 2D array of integers, don\u0026rsquo;t print rows that contain -1 Index at which it becomes greater: Given N and P, form a circle of all the primes under N (including) and start from 2 and keep adding values in clockwise manner till the sum becomes greater than P, return the index at which the sum crosses P.","title":""},{"content":"Thorogood Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 75% XII 75% UG 7.5 GPA Round 1 16/09/23\n3 Sections, no coding\nCommunication Section Aptitude (5) About Thorogood ","permalink":"https://manasch.github.io/placements/code/thorogood/","summary":"Thorogood Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 75% XII 75% UG 7.5 GPA Round 1 16/09/23\n3 Sections, no coding\nCommunication Section Aptitude (5) About Thorogood ","title":""},{"content":"Truminds Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 80% XII 80% UG 8 GPA 45 People with 9+ were shortlisted Bonus\nJoining Bonus - 1,00,000\nRetention Bonus - 4,00,000\nRound 1 - Coding Round 10/09/23\n18 MCQs and 1 coding. Questions were different for everyone.\nCoding Questions String selection: Given N strings where each string is given in the form of an array int of size 26, where the h\u0026rsquo;th element in the string array denotes the count of #h character of lowercase English alphabets in the string. You have to select exactly 4 strings and insert the strings into a Trie. You are allowed to shuffle the characters in each string. Find the minimum number of nodes present in Trie if the strings are selected optimally. Round 2 - Technical Interview Completely based on Resume. Was asked to explain projects and asked basic questions on core subjects. Interview was easy.\nGave 2 approaches for solving anagrams problem and asked which has less time complexity. One approach used 2 for loops (n^2) and other was based on sorting (nlogn). Round 3 - Interview with CEO I was asked questions on various topics in computer science. Questions were not hard. Some people were only asked hr questions.\n","permalink":"https://manasch.github.io/placements/code/truminds/","summary":"Truminds Details Job Status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X 80% XII 80% UG 8 GPA 45 People with 9+ were shortlisted Bonus\nJoining Bonus - 1,00,000\nRetention Bonus - 4,00,000\nRound 1 - Coding Round 10/09/23\n18 MCQs and 1 coding. Questions were different for everyone.\nCoding Questions String selection: Given N strings where each string is given in the form of an array int of size 26, where the h\u0026rsquo;th element in the string array denotes the count of #h character of lowercase English alphabets in the string.","title":""},{"content":"Urban Company Details Job status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 8 GPA OA 31/07/23\nQuestions Black and White tree: Given a tree with N vertices labeled 1-N rooted at 1. The tree is an undirected graph with N-1 edges. Each edge i connects two vertices ai and bi. Each vertex i is colored either white (0) or black (1).\nThe beauty of a vertex i is the number of paths in its subtree that have end vertices of the opposite color.\nFind the beauty of all N vertices.\nNote: A subtree of a vertex i is a connected sub-graph consisting of all the descendants of i including i. Perform dfs and get number of blacks and whites at each vertex and multiply them and store in dp.\n#include\u0026lt;bits/stdc++.h\u0026gt; using namespace std; vector\u0026lt;int\u0026gt; dfs(int i, vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt;\u0026amp; graph, unordered_set\u0026lt;int\u0026gt;\u0026amp; visited, vector\u0026lt;int\u0026gt;\u0026amp; result) { if (visited.find(i) != visited.end()) { return {0, 0}; } visited.insert(i); int black = 0, white = 0; vector\u0026lt;int\u0026gt; numC; for (auto j: graph[i]) { numC = dfs(j, graph, visited, result); black += numC[0]; white += numC[1]; } string[i - 1] == \u0026#39;1\u0026#39; ? ++black : ++white; result[i - 1] = black * white; return {black, white}; } vector\u0026lt;int\u0026gt; solve(int N, string Color, vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt; Edges) { vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt; graph(N + 1); unordered_set\u0026lt;int\u0026gt; visited; vector\u0026lt;int\u0026gt; result(N); for (auto edge: Edges) { graph[edge[0]].push_back(edge[1]); graph[edge[1]].push_back(edge[0]); } dfs(1, graph, visited, result); return result; } This TLE\u0026rsquo;d by 0.005s for 3 cases while the other 7 passed.\nMinimum Value Falling Apples ","permalink":"https://manasch.github.io/placements/code/urban_company/","summary":"Urban Company Details Job status Full Time (Employment + Internship Mandatory)\nCriteria Study Cutoff X % XII % UG 8 GPA OA 31/07/23\nQuestions Black and White tree: Given a tree with N vertices labeled 1-N rooted at 1. The tree is an undirected graph with N-1 edges. Each edge i connects two vertices ai and bi. Each vertex i is colored either white (0) or black (1).\nThe beauty of a vertex i is the number of paths in its subtree that have end vertices of the opposite color.","title":""},{"content":"Walmart Details Job Status Internship Only\nCriteria Study Cutoff X % XII % UG 7 GPA Round 1 15/11/23\n2 Sections\nMCQ (Technical) Coding (2) Coding Questions Mode Selection: You are given the scores of n students, a company wants to select students who have scored the mode of the n students. Return the number of students to reject.\nExample:\n4 2 2 3 4 Output: 2 The mode is 2 and hence the other students who did not score 2 are rejected 6 3 4 3 4 5 6 Output: 4 The mode is either 3 or 4, hence you can reject (4 4 5 6) or (3 3 5 6) making it 4 in total. int solve(vector\u0026lt;int\u0026gt;\u0026amp; scores) { int n = scores.size(); sort(scores.begin(), scored.end()); int mode = 1; int current = 1; for (int i = 1; i \u0026lt; n; ++i) { if (scores[i] == scores[i - 1]) { ++current; } else { mode = max(mode, current); current = 1; } } mode = max(mode, current); return n - mode; } Array By 5: Given an nxm array, replace all the elements in it with the closest number that is divisible by 5. void solve(vector\u0026lt;vector\u0026lt;int\u0026gt;\u0026gt;\u0026amp; arr) { int n = arr.size(); int m = arr[0].size(); for (int i = 0; i \u0026lt; n; ++i) { for (int j = 0; j \u0026lt; n; ++j) { int num = arr[i][j]; int q = num / 5; int lower = 5 * q; int higher = lower + 5; bool isLower = (num - lower) \u0026lt; (higher - num) ? true : false; arr[i][j] = isLower ? lower : higher; } } } ","permalink":"https://manasch.github.io/placements/code/walmart/","summary":"Walmart Details Job Status Internship Only\nCriteria Study Cutoff X % XII % UG 7 GPA Round 1 15/11/23\n2 Sections\nMCQ (Technical) Coding (2) Coding Questions Mode Selection: You are given the scores of n students, a company wants to select students who have scored the mode of the n students. Return the number of students to reject.\nExample:\n4 2 2 3 4 Output: 2 The mode is 2 and hence the other students who did not score 2 are rejected 6 3 4 3 4 5 6 Output: 4 The mode is either 3 or 4, hence you can reject (4 4 5 6) or (3 3 5 6) making it 4 in total.","title":""},{"content":"Zynga Details Job Status Internship Only\nCriteria Study Cutoff X % XII % UG GPA Round 1 13/10/23\nCoding Questions Longest Increasing Subsequence Decode Ways Polynomial Addition ","permalink":"https://manasch.github.io/placements/code/zynga/","summary":"Zynga Details Job Status Internship Only\nCriteria Study Cutoff X % XII % UG GPA Round 1 13/10/23\nCoding Questions Longest Increasing Subsequence Decode Ways Polynomial Addition ","title":""},{"content":"July, 2023 Mon Tue Wed Thu Fri Sat Sun - - - - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 - - - - - - August, 2023 Mon Tue Wed Thu Fri Sat Sun - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 - - - September, 2023 Mon Tue Wed Thu Fri Sat Sun - - - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 - October, 2023 Mon Tue Wed Thu Fri Sat Sun - - - - - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 - - - - - November, 2023 Mon Tue Wed Thu Fri Sat Sun - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 - - - December, 2023 Mon Tue Wed Thu Fri Sat Sun - - - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 January, 2024 Mon Tue Wed Thu Fri Sat Sun 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 - - - - February, 2024 Mon Tue Wed Thu Fri Sat Sun - - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 - - - March, 2024 Mon Tue Wed Thu Fri Sat Sun - - - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 April, 2024 Mon Tue Wed Thu Fri Sat Sun 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 - - - - - May, 2024 Mon Tue Wed Thu Fri Sat Sun - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 - - ","permalink":"https://manasch.github.io/placements/calendar/","summary":"July, 2023 Mon Tue Wed Thu Fri Sat Sun - - - - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 - - - - - - August, 2023 Mon Tue Wed Thu Fri Sat Sun - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 - - - September, 2023 Mon Tue Wed Thu Fri Sat Sun - - - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 - October, 2023 Mon Tue Wed Thu Fri Sat Sun - - - - - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 - - - - - November, 2023 Mon Tue Wed Thu Fri Sat Sun - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 - - - December, 2023 Mon Tue Wed Thu Fri Sat Sun - - - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 January, 2024 Mon Tue Wed Thu Fri Sat Sun 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 - - - - February, 2024 Mon Tue Wed Thu Fri Sat Sun - - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 - - - March, 2024 Mon Tue Wed Thu Fri Sat Sun - - - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 April, 2024 Mon Tue Wed Thu Fri Sat Sun 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 - - - - - May, 2024 Mon Tue Wed Thu Fri Sat Sun - - 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 - - ","title":"Calendar"},{"content":"May, 2024 10-05-2024, Friday Interview Coinswitch - Off Campus: - Process: Offline - Time: 09:00 Add to Google Calendar 07-05-2024, Tuesday Test Coinswitch - Off Campus: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar 05-05-2024, Sunday Register Coinswitch - Off Campus: - Tier: Not Mentioned - Offer: FTE - Deadline: 09:00 Add to Google Calendar March, 2024 04-03-2024, Monday Interview ZScaler: - Process: Online - Time: 10:30 Add to Google Calendar February, 2024 03-02-2024, Saturday Test ZScaler: - Test Mode: Online - Test Time: 11:30 Add to Google Calendar January, 2024 24-01-2024, Wednesday Register Nike ITC: - Tier: Not Mentioned - Offer: Internship - Deadline: 11:00 Add to Google Calendar 17-01-2024, Wednesday Register ZScaler: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar 12-01-2024, Friday Test Anko: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar 11-01-2024, Thursday Register AB InBev GCC: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar 10-01-2024, Wednesday Register Bizom: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: EOD Add to Google Calendar 09-01-2024, Tuesday Register Anko: - Tier: Not Mentioned - Offer: Internship - Deadline: 16:00 Add to Google Calendar 06-01-2024, Saturday Test Rakuten India: - Test Mode: Online - Test Time: 10:30 Add to Google Calendar December, 2023 29-12-2023, Friday Register Rakuten India: - Tier: Not Mentioned - Offer: Internship - Deadline: 09:30 Add to Google Calendar 20-12-2023, Wednesday Test Consoilio LLC: - Test Mode: Offline - Test Time: Not Mentioned Add to Google Calendar Interview Consoilio LLC: - Process: Offline - Time: Not Mentioned Add to Google Calendar 18-12-2023, Monday Test CMTI: - Test Mode: Not Mentioned - Test Time: Not Mentioned Add to Google Calendar 15-12-2023, Friday Register Victoria Secrete: - Tier: Not Mentioned - Offer: Internship - Deadline: 10:00 Add to Google Calendar 14-12-2023, Thursday Test Nextuple: - Test Mode: Offline - Test Time: 12:30 Add to Google Calendar Interview Cohesity: - Process: Online - Time: Not Mentioned Add to Google Calendar Nextuple: - Process: Offline - Time: 10:00 Add to Google Calendar 13-12-2023, Wednesday Interview Oracle PDO: - Process: Offline - Time: Not Mentioned Add to Google Calendar 12-12-2023, Tuesday Register Palo Alto: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test F5 Innovation: - Test Mode: Offline - Test Time: 10:15 Add to Google Calendar Interview F5 Innovation: - Process: Offline - Time: 13:30 Add to Google Calendar 11-12-2023, Monday Register MediBuddy: - Tier: Not Mentioned - Offer: Internship - Deadline: 10:00 Add to Google Calendar Test Cohesity: - Test Mode: Online - Test Time: 18:00 Add to Google Calendar Netcore Cloud: - Test Mode: Offline - Test Time: 09:30 Add to Google Calendar Interview Netcore Cloud: - Process: Offline - Time: Not Mentioned Add to Google Calendar Target: - Process: Online - Time: Not Mentioned Add to Google Calendar 09-12-2023, Saturday Register Cubic Logics: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar 08-12-2023, Friday Register AtkinsRealis: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Cohesity: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Netcore Cloud: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar IQVUA: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 09:00 Add to Google Calendar Test Conga: - Test Mode: Online - Test Time: 19:00 Add to Google Calendar Newstreet Tech: - Test Mode: Online - Test Time: 20:00 Add to Google Calendar 07-12-2023, Thursday Register Oracle PDO: - Tier: Not Mentioned - Offer: FTE - Deadline: 09:00 Add to Google Calendar JAR APP: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Metric: - Tier: Not Mentioned - Offer: Internship - Deadline: 09:00 Add to Google Calendar Test Oracle PDO: - Test Mode: Online - Test Time: 19:00 Add to Google Calendar Interview Akami Technologies: - Process: Offline - Time: Not Mentioned Add to Google Calendar 05-12-2023, Tuesday Register Biocliq AI: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: EOD Add to Google Calendar Flatworld: - Tier: Not Mentioned - Offer: Internship - Deadline: 09:00 Add to Google Calendar MatreComm Tech.: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 09:00 Add to Google Calendar Conga: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test Target: - Test Mode: Online - Test Time: 19:00 Add to Google Calendar Akami Technologies: - Test Mode: Offline - Test Time: 21:00 Add to Google Calendar 04-12-2023, Monday Interview Observe.ai: - Process: Offline - Time: Not Mentioned Add to Google Calendar 03-12-2023, Sunday Register Target: - Tier: Not Mentioned - Offer: Internship - Deadline: EOD Add to Google Calendar Test Groww: - Test Mode: Online - Test Time: 19:00 Add to Google Calendar 02-12-2023, Saturday Register Rattle: - Tier: Not Mentioned - Offer: Internship - Deadline: EOD Add to Google Calendar Triomics: - Tier: Not Mentioned - Offer: Internship - Deadline: EOD Add to Google Calendar 01-12-2023, Friday Register Publicis Sapient: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar November, 2023 30-11-2023, Thursday Register Philips: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 09:00 Add to Google Calendar Observe.ai: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Newstreet Tech: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 09:30 Add to Google Calendar Microland: - Tier: Not Mentioned - Offer: Not Mentioned - Deadline: 09:00 Add to Google Calendar Test Observe.ai: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar 29-11-2023, Wednesday Register Aptos India: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 10:00 Add to Google Calendar 28-11-2023, Tuesday Register Akami Technologies: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:30 Add to Google Calendar Test phData Solutions: - Test Mode: Offline - Test Time: Not Mentioned Add to Google Calendar Interview phData Solutions: - Process: Offline - Time: Not Mentioned Add to Google Calendar Berkadia: - Process: Online - Time: Not Mentioned Add to Google Calendar 27-11-2023, Monday Register Astr Defence: - Tier: Not Mentioned - Offer: Internship - Deadline: EOD Add to Google Calendar Test Deltax: - Test Mode: Online - Test Time: 12:30 Add to Google Calendar Interview MIQ Digital: - Process: Offline - Time: 09:30 Add to Google Calendar Deltax: - Process: Online - Time: 18:00 Add to Google Calendar 25-11-2023, Saturday Register F5 Innovation: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Interview D.E. Shaw: - Process: Offline - Time: 09:00 Add to Google Calendar 23-11-2023, Thursday Register Reltio: - Tier: Not Mentioned - Offer: FTE - Deadline: 09:00 Add to Google Calendar Interview Oracle: - Process: Offline - Time: Not Mentioned Add to Google Calendar 22-11-2023, Wednesday Register Quicken Inc.: - Tier: Not Mentioned - Offer: Internship - Deadline: 11:00 Add to Google Calendar phData Solutions: - Tier: Not Mentioned - Offer: FTE - Deadline: 09:00 Add to Google Calendar Interview Netradyne Tech: - Process: Offline - Time: 09:00 Add to Google Calendar 21-11-2023, Tuesday Register Oracle: - Tier: Not Mentioned - Offer: Not Mentioned - Deadline: 08:00 Add to Google Calendar Test Oracle: - Test Mode: Online - Test Time: 17:00 Add to Google Calendar Netradyne Tech: - Test Mode: Offline - Test Time: 08:00 Add to Google Calendar Interview IDFC First Bank: - Process: Offline - Time: 10:00 Add to Google Calendar 20-11-2023, Monday Test D.E. Shaw: - Test Mode: Online - Test Time: 11:00 Add to Google Calendar 18-11-2023, Saturday Register Consoilio LLC: - Tier: Not Mentioned - Offer: FTE - Deadline: EOD Add to Google Calendar Q2: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 09:00 Add to Google Calendar Digicert: - Tier: Not Mentioned - Offer: Internship - Deadline: 09:00 Add to Google Calendar Netradyne Tech: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar 17-11-2023, Friday Register Groww: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 09:00 Add to Google Calendar Test IDFC First Bank: - Test Mode: Online - Test Time: 18:00 Add to Google Calendar 16-11-2023, Thursday Register Udaan.com: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 08:00 Add to Google Calendar Test AT\u0026amp;T: - Test Mode: Offline - Test Time: 17:00 Add to Google Calendar 15-11-2023, Wednesday Test Walmart: - Test Mode: Online - Test Time: 18:00 Add to Google Calendar Inflobox: - Test Mode: Offline - Test Time: Not Mentioned Add to Google Calendar Interview Inflobox: - Process: Offline - Time: Not Mentioned Add to Google Calendar 14-11-2023, Tuesday Register IDFC First Bank: - Tier: Not Mentioned - Offer: FTE - Deadline: 09:00 Add to Google Calendar 13-11-2023, Monday Register Walmart: - Tier: Not Mentioned - Offer: Internship - Deadline: 22:30 Add to Google Calendar Test GSK GCC: - Test Mode: Online - Test Time: 23:55 Add to Google Calendar Gyansys: - Test Mode: Offline - Test Time: Not Mentioned Add to Google Calendar Interview Gyansys: - Process: Offline - Time: Not Mentioned Add to Google Calendar 11-11-2023, Saturday Register Cubic Logics: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar JLL Technology: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Test MIQ Digital: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar 10-11-2023, Friday Register Alcon India: - Tier: Not Mentioned - Offer: Internship - Deadline: 09:30 Add to Google Calendar 09-11-2023, Thursday Register GSK GCC: - Tier: Not Mentioned - Offer: Internship - Deadline: 09:30 Add to Google Calendar Drongo AI: - Tier: Not Mentioned - Offer: Internship - Deadline: EOD Add to Google Calendar Test Berkadia: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar 08-11-2023, Wednesday Register D.E. Shaw: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 14:00 Add to Google Calendar BNP Paribas: - Tier: Not Mentioned - Offer: Internship - Deadline: 09:00 Add to Google Calendar Gyansys: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 10:30 Add to Google Calendar Test Shadowfax: - Test Mode: Offline - Test Time: 10:00 Add to Google Calendar Interview Shadowfax: - Process: Offline - Time: 12:00 Add to Google Calendar Microchip (Again): - Process: Offline - Time: Not Mentioned Add to Google Calendar 07-11-2023, Tuesday Register MIQ Digital: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 19:30 Add to Google Calendar Test Keysight: - Test Mode: Online - Test Time: 17:00 Add to Google Calendar Microchip (Again): - Test Mode: Online - Test Time: 07:30 Add to Google Calendar Interview Deloitte India Consulting: - Process: Online - Time: 10:15 Add to Google Calendar Aurigo: - Process: Offline - Time: Not Mentioned Add to Google Calendar Endor Labs: - Process: Offline - Time: 09:00 Add to Google Calendar 06-11-2023, Monday Interview Garrett Motion: - Process: Offline - Time: 09:00 Add to Google Calendar 05-11-2023, Sunday Register Shadowfax: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar 04-11-2023, Saturday Register Keysight: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 09:00 Add to Google Calendar Celstream: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Test Aurigo: - Test Mode: Online - Test Time: 10:00 Add to Google Calendar Endor Labs: - Test Mode: Online - Test Time: 11:00 Add to Google Calendar 03-11-2023, Friday Test Garrett Motion: - Test Mode: Online - Test Time: 12:00 Add to Google Calendar Interview Mercedes-Benz R\u0026amp;D: - Process: Offline - Time: Not Mentioned Add to Google Calendar 02-11-2023, Thursday Test Deloitte India Consulting: - Test Mode: Online - Test Time: 16:00 Add to Google Calendar Interview Intel: - Process: Offline - Time: 09:00 Add to Google Calendar 01-11-2023, Wednesday Register Inflobox: - Tier: Not Mentioned - Offer: Internship - Deadline: EOD Add to Google Calendar Deloitte India Consulting: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 09:00 Add to Google Calendar Aurigo: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Garrett Motion: - Tier: Not Mentioned - Offer: Internship - Deadline: 10:00 Add to Google Calendar Test Paypal: - Test Mode: Online - Test Time: 18:00 Add to Google Calendar October, 2023 31-10-2023, Tuesday Register Microchip (Again): - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Cyware Labs: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: EOD Add to Google Calendar Interview BlueYonder: - Process: Offline - Time: 09:00 Add to Google Calendar Juniper Networks: - Process: Online - Time: 09:00 Add to Google Calendar Nvidia: - Process: Offline - Time: 09:00 Add to Google Calendar 30-10-2023, Monday Test BlueYonder: - Test Mode: Online - Test Time: 13:30 Add to Google Calendar Interview Zebra Tech: - Process: Offline - Time: 09:00 Add to Google Calendar 29-10-2023, Sunday Register Conde Nast: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar 27-10-2023, Friday Register BlueYonder: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test Mercedes-Benz R\u0026amp;D: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar Interview IBM Consulting: - Process: Offline - Time: 09:00 Add to Google Calendar EY India: - Process: Offline - Time: 08:30 Add to Google Calendar 26-10-2023, Thursday Register Endor Labs: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 09:00 Add to Google Calendar Berkadia: - Tier: Not Mentioned - Offer: FTE - Deadline: 09:00 Add to Google Calendar Havells India: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 10:00 Add to Google Calendar Test IBM Consulting: - Test Mode: Offline - Test Time: 11:30 Add to Google Calendar Juniper Networks: - Test Mode: Online - Test Time: 19:00 Add to Google Calendar 24-10-2023, Tuesday Register Mercedes-Benz R\u0026amp;D: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 10:30 Add to Google Calendar AT\u0026amp;T: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 10:00 Add to Google Calendar 22-10-2023, Sunday Register C-DOT: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar 20-10-2023, Friday Register Intel: - Tier: Not Mentioned - Offer: Internship - Deadline: 09:00 Add to Google Calendar Test Zebra Tech: - Test Mode: Offline - Test Time: 09:30 Add to Google Calendar Interview AstraZeneca: - Process: Offline - Time: 09:00 Add to Google Calendar Subex: - Process: Offline - Time: 09:30 Add to Google Calendar 19-10-2023, Thursday Register Nextuple: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar CRK Gida: - Tier: Not Mentioned - Offer: FTE - Deadline: EOD Add to Google Calendar Zebra Tech: - Tier: Not Mentioned - Offer: Internship - Deadline: 08:00 Add to Google Calendar 18-10-2023, Wednesday Test AstraZeneca: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar Samsung Research Institute: - Test Mode: Online - Test Time: 19:00 Add to Google Calendar Interview IDFY: - Process: Offline - Time: 09:00 Add to Google Calendar Microchip: - Process: Offline - Time: Not Mentioned Add to Google Calendar 17-10-2023, Tuesday Register AstraZeneca: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test Microchip: - Test Mode: Online - Test Time: 07:00 Add to Google Calendar Interview Zynga: - Process: Offline - Time: Not Mentioned Add to Google Calendar 16-10-2023, Monday Register IDFY: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Test IDFY: - Test Mode: Online - Test Time: 18:00 Add to Google Calendar Perfios: - Test Mode: Offline - Test Time: 09:30 Add to Google Calendar Interview Perfios: - Process: Offline - Time: 11:30 Add to Google Calendar 14-10-2023, Saturday Register Samsung Research Institute: - Tier: Not Mentioned - Offer: Internship - Deadline: EOD Add to Google Calendar Microchip: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 16:30 Add to Google Calendar Test Aurva.io: - Test Mode: Online - Test Time: 17:00 Add to Google Calendar 13-10-2023, Friday Test Zynga: - Test Mode: Online - Test Time: 17:30 Add to Google Calendar 12-10-2023, Thursday Register Aurva.io: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 10:00 Add to Google Calendar Perfios: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 10:30 Add to Google Calendar Interview Tejas Networks: - Process: Offline - Time: 09:00 Add to Google Calendar 11-10-2023, Wednesday Register Juniper Networks: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: EOD Add to Google Calendar Collins Aerospace: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 10:00 Add to Google Calendar Interview Infibeam: - Process: Offline - Time: 10:00 Add to Google Calendar 10-10-2023, Tuesday Test Tejas Networks: - Test Mode: Offline - Test Time: 14:00 Add to Google Calendar 09-10-2023, Monday Register Zynga: - Tier: Not Mentioned - Offer: Internship - Deadline: 17:00 Add to Google Calendar Apple: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 13:00 Add to Google Calendar Test Confluent: - Test Mode: Offline - Test Time: Not Mentioned Add to Google Calendar Interview Confluent: - Process: Offline - Time: Not Mentioned Add to Google Calendar 08-10-2023, Sunday Test EY India: - Test Mode: Online - Test Time: 19:00 Add to Google Calendar 07-10-2023, Saturday Test Infibeam: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar Hiver: - Test Mode: Online - Test Time: 19:00 Add to Google Calendar Interview Pratt \u0026amp; Whitney: - Process: Offline - Time: 09:00 Add to Google Calendar 06-10-2023, Friday Register IBM Consulting: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 11:00 Add to Google Calendar Test Subex: - Test Mode: Online - Test Time: 20:00 Add to Google Calendar Interview Samsung Semiconductors: - Process: Online - Time: Not Mentioned Add to Google Calendar Amadeus: - Process: Offline - Time: 09:00 Add to Google Calendar 05-10-2023, Thursday Register Samsung Semiconductors: - Tier: Not Mentioned - Offer: Not Mentioned - Deadline: 08:30 Add to Google Calendar Infibeam: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Aveva: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Tejas Networks: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 10:00 Add to Google Calendar Test Samsung Semiconductors: - Test Mode: Online - Test Time: 16:00 Add to Google Calendar Pratt \u0026amp; Whitney: - Test Mode: Online - Test Time: 19:00 Add to Google Calendar Amadeus: - Test Mode: Offline - Test Time: 09:00 Add to Google Calendar Interview KPMG: - Process: Offline - Time: Not Mentioned Add to Google Calendar 04-10-2023, Wednesday Register Hiver: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Test KPMG: - Test Mode: Offline - Test Time: Not Mentioned Add to Google Calendar 03-10-2023, Tuesday Register Pratt \u0026amp; Whitney: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 10:30 Add to Google Calendar 01-10-2023, Sunday Test ARM: - Test Mode: Online - Test Time: 10:00 Add to Google Calendar September, 2023 30-09-2023, Saturday Register Siemens: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar 28-09-2023, Thursday Register Confluent: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: EOD Add to Google Calendar 27-09-2023, Wednesday Register Deltax: - Tier: Not Mentioned - Offer: FTE - Deadline: 09:00 Add to Google Calendar Interview Dover: - Process: Online - Time: 10:00 Add to Google Calendar Goldman Sachs: - Process: Offline - Time: Not Mentioned Add to Google Calendar 26-09-2023, Tuesday Register KPMG: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar 25-09-2023, Monday Register Paypal: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Test Dover: - Test Mode: Online - Test Time: 18:00 Add to Google Calendar Interview Broadridge: - Process: Offline - Time: 09:00 Add to Google Calendar Arup India Pvt. Ltd.: - Process: Offline - Time: Not Mentioned Add to Google Calendar 23-09-2023, Saturday Test SAP Labs: - Test Mode: Online - Test Time: 17:00 Add to Google Calendar 22-09-2023, Friday Register ARM: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Interview British Telecom: - Process: Offline - Time: 09:00 Add to Google Calendar 21-09-2023, Thursday Register SecPod: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 09:00 Add to Google Calendar Dover: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar SAP Labs: - Tier: Not Mentioned - Offer: FTE - Deadline: 09:00 Add to Google Calendar Test Arup India Pvt. Ltd.: - Test Mode: Offline - Test Time: 09:00 Add to Google Calendar Interview Morgan Stanley: - Process: Offline - Time: 09:00 Add to Google Calendar 20-09-2023, Wednesday Test Nvidia: - Test Mode: Online - Test Time: 19:30 Add to Google Calendar Interview Thorogood: - Process: Offline - Time: 09:00 Add to Google Calendar 19-09-2023, Tuesday Test British Telecom: - Test Mode: Online - Test Time: 17:00 Add to Google Calendar 17-09-2023, Sunday Register Subex: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: EOD Add to Google Calendar Test Morgan Stanley: - Test Mode: Online - Test Time: 11:00 Add to Google Calendar 16-09-2023, Saturday Register Adobe: - Tier: Not Mentioned - Offer: Internship - Deadline: 09:00 Add to Google Calendar British Telecom: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test Goldman Sachs: - Test Mode: Online - Test Time: 10:00 Add to Google Calendar MakeMyTrip: - Test Mode: Online - Test Time: 11:00 Add to Google Calendar Thorogood: - Test Mode: Online - Test Time: 17:00 Add to Google Calendar 15-09-2023, Friday Register Goldman Sachs: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 09:00 Add to Google Calendar Test Broadridge: - Test Mode: Online - Test Time: 09:00 Add to Google Calendar Interview Eli Lilly: - Process: Offline - Time: 09:00 Add to Google Calendar IBM: - Process: Offline - Time: Not Mentioned Add to Google Calendar Lam Research: - Process: Offline - Time: Not Mentioned Add to Google Calendar 14-09-2023, Thursday Register Nvidia: - Tier: Not Mentioned - Offer: Internship - Deadline: 09:30 Add to Google Calendar Morgan Stanley: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar MakeMyTrip: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 09:00 Add to Google Calendar Test Eli Lilly: - Test Mode: Online - Test Time: 10:00 Add to Google Calendar Interview Bain Capability: - Process: Online - Time: 09:30 Add to Google Calendar 13-09-2023, Wednesday Register Thorogood: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test Lam Research: - Test Mode: Online - Test Time: 18:00 Add to Google Calendar Interview Trueminds: - Process: Offline - Time: 09:00 Add to Google Calendar 12-09-2023, Tuesday Register Broadridge: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Eli Lilly: - Tier: Not Mentioned - Offer: FTE - Deadline: 09:00 Add to Google Calendar Interview Societe Generale: - Process: Offline - Time: Not Mentioned Add to Google Calendar 11-09-2023, Monday Interview Deloitte: - Process: Offline - Time: Not Mentioned Add to Google Calendar 10-09-2023, Sunday Test IBM: - Test Mode: Online - Test Time: 10:00 Add to Google Calendar Trueminds: - Test Mode: Online - Test Time: 17:00 Add to Google Calendar 09-09-2023, Saturday Test Bain Capability: - Test Mode: Online - Test Time: 12:00 Add to Google Calendar 08-09-2023, Friday Test Accenture: - Test Mode: Offline - Test Time: 09:00 Add to Google Calendar Interview HPE: - Process: Online - Time: 09:00 Add to Google Calendar Natwest: - Process: Offline - Time: Not Mentioned Add to Google Calendar 07-09-2023, Thursday Register Arup India Pvt. Ltd.: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Tech Mahindra: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar IBM: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar 06-09-2023, Wednesday Test Societe Generale: - Test Mode: Online - Test Time: 18:00 Add to Google Calendar HPE: - Test Mode: Online - Test Time: 14:00 Add to Google Calendar Interview Schneider Electric: - Process: Offline - Time: Not Mentioned Add to Google Calendar 05-09-2023, Tuesday Register Societe Generale: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 17:00 Add to Google Calendar Trueminds: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar 04-09-2023, Monday Interview Sprinklr: - Process: Offline - Time: 10:00 Add to Google Calendar 03-09-2023, Sunday Register Bain Capability: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Test Deloitte: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar 02-09-2023, Saturday Register Deloitte: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Test Sprinklr: - Test Mode: Online - Test Time: 11:00 Add to Google Calendar 01-09-2023, Friday Test DXC: - Test Mode: Online - Test Time: 12:00 Add to Google Calendar Interview Cisco: - Process: Online - Time: 08:00 Add to Google Calendar August, 2023 31-08-2023, Thursday Register EY India: - Tier: Not Mentioned - Offer: Not Mentioned - Deadline: 10:30 Add to Google Calendar DXC: - Tier: Not Mentioned - Offer: FTE - Deadline: 09:00 Add to Google Calendar Interview Applied Materials: - Process: Offline - Time: Not Mentioned Add to Google Calendar Volvo: - Process: Offline - Time: Not Mentioned Add to Google Calendar 30-08-2023, Wednesday Register Sprinklr: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test Dish: - Test Mode: Offline - Test Time: 09:00 Add to Google Calendar Interview Dish: - Process: Offline - Time: 13:00 Add to Google Calendar 29-08-2023, Tuesday Register Applied Materials: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 11:00 Add to Google Calendar Dish: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Amadeus: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 08:30 Add to Google Calendar Test Applied Materials: - Test Mode: Online - Test Time: 19:00 Add to Google Calendar Natwest: - Test Mode: Offline - Test Time: 15:00 Add to Google Calendar Cisco: - Test Mode: Offline - Test Time: 08:00 Add to Google Calendar Interview Kickdrum: - Process: Offline - Time: Not Mentioned Add to Google Calendar 28-08-2023, Monday Test Schneider Electric: - Test Mode: Online - Test Time: 16:00 Add to Google Calendar Interview Epsilon: - Process: Offline - Time: 09:00 Add to Google Calendar 27-08-2023, Sunday Test CynLr: - Test Mode: Online - Test Time: 10:00 Add to Google Calendar 26-08-2023, Saturday Register HPE: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Test Volvo: - Test Mode: Offline - Test Time: Not Mentioned Add to Google Calendar 25-08-2023, Friday Register Schneider Electric: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 08:30 Add to Google Calendar Volvo: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Test AMETEK: - Test Mode: Offline - Test Time: Not Mentioned Add to Google Calendar Kickdrum: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar Baxter: - Test Mode: Offline - Test Time: Not Mentioned Add to Google Calendar Interview AMETEK: - Process: Offline - Time: Not Mentioned Add to Google Calendar Baxter: - Process: Offline - Time: Not Mentioned Add to Google Calendar 24-08-2023, Thursday Register AMETEK: - Tier: Not Mentioned - Offer: Internship - Deadline: EOD Add to Google Calendar Test Epsilon: - Test Mode: Offline - Test Time: 15:00 Add to Google Calendar Sahaj: - Test Mode: Offline - Test Time: Not Mentioned Add to Google Calendar Interview Sahaj: - Process: Offline - Time: Not Mentioned Add to Google Calendar Hyperface Tech: - Process: Offline - Time: Not Mentioned Add to Google Calendar 23-08-2023, Wednesday Register Natwest: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar CynLr: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test Newfold Digital: - Test Mode: Online - Test Time: 08:00 Add to Google Calendar Interview Cloudera: - Process: Offline - Time: 09:00 Add to Google Calendar 22-08-2023, Tuesday Register Cisco: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 17:00 Add to Google Calendar Epsilon: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test Sandvine: - Test Mode: Offline - Test Time: 08:30 Add to Google Calendar Interview Sandvine: - Process: Offline - Time: 14:00 Add to Google Calendar 21-08-2023, Monday Register Kickdrum: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test Cloudera: - Test Mode: Online - Test Time: 17:00 Add to Google Calendar Interview Egnyte: - Process: Offline - Time: Not Mentioned Add to Google Calendar 19-08-2023, Saturday Test Egnyte: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar 18-08-2023, Friday Register Baxter: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 12:00 Add to Google Calendar Sahaj: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 12:00 Add to Google Calendar Test Hyperface Tech: - Test Mode: Offline - Test Time: 09:00 Add to Google Calendar 17-08-2023, Thursday Test Arista: - Test Mode: Offline - Test Time: 09:00 Add to Google Calendar Interview Arista: - Process: Offline - Time: 11:00 Add to Google Calendar 16-08-2023, Wednesday Interview Deutsche: - Process: Offline - Time: 09:00 Add to Google Calendar 15-08-2023, Tuesday Register Cloudera: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar 13-08-2023, Sunday Register Egnyte: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar 11-08-2023, Friday Register Oracle: - Tier: Not Mentioned - Offer: FTE - Deadline: 13:00 Add to Google Calendar Arista: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test Deutsche: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar Interview Tesco: - Process: Offline - Time: 09:15 Add to Google Calendar 10-08-2023, Thursday Register Tesco: - Tier: Not Mentioned - Offer: FTE - Deadline: 09:00 Add to Google Calendar Test Tesco: - Test Mode: Online - Test Time: 21:00 Add to Google Calendar Interview Swym: - Process: Offline - Time: 10:00 Add to Google Calendar 09-08-2023, Wednesday Register Newfold Digital: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 14:00 Add to Google Calendar Hyperface Tech: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test Swym: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar Couchbase: - Test Mode: Not Mentioned - Test Time: Not Mentioned Add to Google Calendar Interview Couchbase: - Process: Offline - Time: 08:00 Add to Google Calendar 08-08-2023, Tuesday Register Sandvine: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test PWC: - Test Mode: Offline - Test Time: Not Mentioned Add to Google Calendar Interview Ring Central: - Process: Offline - Time: 09:00 Add to Google Calendar 07-08-2023, Monday Interview Pure Storage: - Process: Offline - Time: 09:00 Add to Google Calendar 06-08-2023, Sunday Test Ring Central: - Test Mode: Online - Test Time: 12:00 Add to Google Calendar 05-08-2023, Saturday Register Deutsche: - Tier: Not Mentioned - Offer: FTE - Deadline: 09:00 Add to Google Calendar Swym: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Lam Research: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar 04-08-2023, Friday Register Couchbase: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 20:00 Add to Google Calendar Interview Commvault: - Process: Offline - Time: 09:00 Add to Google Calendar 02-08-2023, Wednesday Register Ring Central: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 20:00 Add to Google Calendar Test Pure Storage: - Test Mode: Online - Test Time: 19:00 Add to Google Calendar Interview Urban Company: - Process: Offline - Time: 09:30 Add to Google Calendar 01-08-2023, Tuesday Register Pure Storage: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test Commvault: - Test Mode: Offline - Test Time: 11:30 Add to Google Calendar July, 2023 31-07-2023, Monday Test Urban Company: - Test Mode: Online - Test Time: 19:00 Add to Google Calendar Interview Intuit: - Process: Offline - Time: 09:00 Add to Google Calendar 29-07-2023, Saturday Register Urban Company: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar 27-07-2023, Thursday Test Intuit: - Test Mode: Online - Test Time: 18:00 Add to Google Calendar 25-07-2023, Tuesday Register Commvault: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 10:30 Add to Google Calendar Intuit: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 01:30 Add to Google Calendar 13-07-2023, Thursday Register Accenture: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar ","permalink":"https://manasch.github.io/placements/schedule/","summary":"May, 2024 10-05-2024, Friday Interview Coinswitch - Off Campus: - Process: Offline - Time: 09:00 Add to Google Calendar 07-05-2024, Tuesday Test Coinswitch - Off Campus: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar 05-05-2024, Sunday Register Coinswitch - Off Campus: - Tier: Not Mentioned - Offer: FTE - Deadline: 09:00 Add to Google Calendar March, 2024 04-03-2024, Monday Interview ZScaler: - Process: Online - Time: 10:30 Add to Google Calendar February, 2024 03-02-2024, Saturday Test ZScaler: - Test Mode: Online - Test Time: 11:30 Add to Google Calendar January, 2024 24-01-2024, Wednesday Register Nike ITC: - Tier: Not Mentioned - Offer: Internship - Deadline: 11:00 Add to Google Calendar 17-01-2024, Wednesday Register ZScaler: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar 12-01-2024, Friday Test Anko: - Test Mode: Online - Test Time: Not Mentioned Add to Google Calendar 11-01-2024, Thursday Register AB InBev GCC: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar 10-01-2024, Wednesday Register Bizom: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: EOD Add to Google Calendar 09-01-2024, Tuesday Register Anko: - Tier: Not Mentioned - Offer: Internship - Deadline: 16:00 Add to Google Calendar 06-01-2024, Saturday Test Rakuten India: - Test Mode: Online - Test Time: 10:30 Add to Google Calendar December, 2023 29-12-2023, Friday Register Rakuten India: - Tier: Not Mentioned - Offer: Internship - Deadline: 09:30 Add to Google Calendar 20-12-2023, Wednesday Test Consoilio LLC: - Test Mode: Offline - Test Time: Not Mentioned Add to Google Calendar Interview Consoilio LLC: - Process: Offline - Time: Not Mentioned Add to Google Calendar 18-12-2023, Monday Test CMTI: - Test Mode: Not Mentioned - Test Time: Not Mentioned Add to Google Calendar 15-12-2023, Friday Register Victoria Secrete: - Tier: Not Mentioned - Offer: Internship - Deadline: 10:00 Add to Google Calendar 14-12-2023, Thursday Test Nextuple: - Test Mode: Offline - Test Time: 12:30 Add to Google Calendar Interview Cohesity: - Process: Online - Time: Not Mentioned Add to Google Calendar Nextuple: - Process: Offline - Time: 10:00 Add to Google Calendar 13-12-2023, Wednesday Interview Oracle PDO: - Process: Offline - Time: Not Mentioned Add to Google Calendar 12-12-2023, Tuesday Register Palo Alto: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Test F5 Innovation: - Test Mode: Offline - Test Time: 10:15 Add to Google Calendar Interview F5 Innovation: - Process: Offline - Time: 13:30 Add to Google Calendar 11-12-2023, Monday Register MediBuddy: - Tier: Not Mentioned - Offer: Internship - Deadline: 10:00 Add to Google Calendar Test Cohesity: - Test Mode: Online - Test Time: 18:00 Add to Google Calendar Netcore Cloud: - Test Mode: Offline - Test Time: 09:30 Add to Google Calendar Interview Netcore Cloud: - Process: Offline - Time: Not Mentioned Add to Google Calendar Target: - Process: Online - Time: Not Mentioned Add to Google Calendar 09-12-2023, Saturday Register Cubic Logics: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar 08-12-2023, Friday Register AtkinsRealis: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: EOD Add to Google Calendar Cohesity: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Netcore Cloud: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar IQVUA: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: 09:00 Add to Google Calendar Test Conga: - Test Mode: Online - Test Time: 19:00 Add to Google Calendar Newstreet Tech: - Test Mode: Online - Test Time: 20:00 Add to Google Calendar 07-12-2023, Thursday Register Oracle PDO: - Tier: Not Mentioned - Offer: FTE - Deadline: 09:00 Add to Google Calendar JAR APP: - Tier: Not Mentioned - Offer: FTE, Internship - Deadline: 09:00 Add to Google Calendar Metric: - Tier: Not Mentioned - Offer: Internship - Deadline: 09:00 Add to Google Calendar Test Oracle PDO: - Test Mode: Online - Test Time: 19:00 Add to Google Calendar Interview Akami Technologies: - Process: Offline - Time: Not Mentioned Add to Google Calendar 05-12-2023, Tuesday Register Biocliq AI: - Tier: Not Mentioned - Offer: Internship, PPO - Deadline: EOD Add to Google Calendar Flatworld: - Tier: Not Mentioned - Offer: Internship - Deadline: 09:00 Add to Google Calendar MatreComm Tech.","title":"Schedule"}]