Morgan Stanley

Details

Job Status

Full Time (Employment + Internship Mandatory)

Criteria

StudyCutoff
X%
XII%
UG8 GPA

Round 1

17/09/23

There were 3 sections and a total of 100 min.

  1. Automata Fix (7) - 20min - (A debugging section, debug code such that it passes all test cases.)
  2. Quantitative and Logical Aptitude (10) - 20min
  3. Coding (3) - 60min

Coding Questions

  1. Longest Common Prefix: Given an array of names, find the longest common prefix of all the names.
string solve(vector<string>& names, int n) {
	if (n == 0) {
		return "";
	}
	if (n == 1) {
		return names[0];
	}
	
	sort(names.begin(), names.end());
	string min_string = names.front().size() < names.back().size() ? names.front() : names.back();
	int min_length = min_string.size();
	
	int i = 0;
	string res = "";
	while (i < min_length && names.front()[i] == names.back()[i]) {
		res += min_string[i];
		++i;
	}
	return res;
}

int main()
{
	int n;
	cin >> n;
	vector<string> names(n);
	for (int i = 0; i < n; ++i) {
		cin >> names[i];
        transform(names[i].begin(), names[i].end(), names[i].begin(), [](unsigned char c) {return tolower(c);});
	}
	
	string res = solve(names, n);
	cout << res << endl;
	return 0;
}
  1. 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 >=0 and r < 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 < n - 1; ++i) {
        l = i;
        r = i + 1;
        leftSum = 0;
        rightSum = 0;
        while (l >= 0 && r < n) {
            leftSum += num[l] - '0';
            rightSum += num[r] - '0';
            if (leftSum == rightSum) {
                maxLen = max(maxLen, r - l + 1);
            }
            --l;
            ++r;
        }
    }
    return maxLen;
}

int main()
{
    string num;
    cin >> num;
    
    int res = solve(num, num.size());
    cout << res << endl;

    return 0;
}

  1. Number of Different Plants: Given a 2D matrix of 0’s and 1’s, find the different number plants that exist. In the cell, if a plant exists, it is marked as 1 otherwise 0.

    The 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.

int solve(vector<vector<int>>& matrix, int m, int n) {
	int count = 0;
	set<pair<int, int>> visited;
	
	auto bfs = [&] (auto self, int i, int j) {
		if (min(i, j) < 0 || i >= m || j >= 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 < m; ++i) {
		for (int j = 0; j < n; ++j) {
			if (matrix[i][j] == 1 && visited.find({i, j}) == visited.end()) {
				bfs(bfs, i, j);
				++count;
			}
		}
	}
	
	return count;
}

int main()
{
	int n, m;
	cin >> n >> m;
	vector<vector<int>> graph(n, vector<int>(m, 0));
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < m; ++j) {
			cin >> graph[i][j];
		}
	}
	
	int ans = solve(graph, n, m);
	cout << ans << endl;
	return 0;
}
  • This is exactly LC#200 but worded differently.