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int solve(vector<vector<int>> arr) {
    int n = arr.size();
    vector<vector<int>> dp(n, vector<int>(n));
    int maxi = 0;

    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            if(i == 0 || j == 0) {
                dp[i][j] = arr[i][j];
            } else if(arr[i][j] == 1) {
                dp[i][j] = min({dp[i-1][j], dp[i][j-1], dp[i-1][j-1]}) + 1;
            }
            maxi = max(maxi, dp[i][j]);
        }
    }

    return maxi;
}

IBM

C++

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return min(e-b+1, numChapters)
Python 3βœ…
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IBMβœ…
SQL

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string make_string_S_to_T(string S) {
    string T=β€œprogramming”;
    bool possible = false;
    int M = T.length();
    int N = S.length();
    for (int i = 0; i <= M; i++) {
        int prefix_length = i;
        int suffix_length = M - i;
        string prefix = S.substr(0, prefix_length);
        string suffix = S.substr(N - suffix_length, suffix_length);
        if (prefix + suffix == T) {
            possible = true;
            break;
        }
    }
    if (possible)
        return "YES";
    else
        return "NO";
}

Deleting substring βœ…
Zeta

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