Determine what is the minimum time required to rot all oranges.Given a matrix of dimension r*c where each cell in the matrix can have values 0, 1, or 2.

                      to determine what is the minimum time required to rot all oranges

Given a matrix of dimension r*c where each cell in the matrix can have values 0, 1, or 2 which has the following meaning:
0: Empty cell
1: Cells have fresh oranges
2: Cells have rotten oranges
So, we have to determine what is the minimum time required to rot all oranges. A rotten orange at index [i,j] can rot other fresh orange at indexes [i-1,j], [i+1,j], [i,j-1], [i,j+1] (up, down, left and right) in unit time. If it is impossible to rot every orange then simply return -1.

Here is a simple and easy solution using queue with execution time is 0.06s.

#include <bits/stdc++.h>

using namespace std;

int solveRotten(vector < vector < int > > a, int n, int m) {
    int timer = -1;
    queue < pair < int, int > > q;

    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            if (a[i][j] == 2) {
                q.push(make_pair(i, j));
            }
        }
    }

    while (!q.empty()) {
        int siz = q.size();

        for (int i = 0; i < siz; ++i) {
            pair < int, int > p = q.front();
            int j = p.first, k = p.second;

            if (j - 1 >= 0 && a[j - 1][k] == 1) {
                q.push(make_pair(j - 1, k));
                a[j - 1][k] = 2;
            }

            if (j + 1 < n && a[j + 1][k] == 1) {
                q.push(make_pair(j + 1, k));
                a[j + 1][k] = 2;
            }

            if (k - 1 >= 0 && a[j][k - 1] == 1) {
                q.push(make_pair(j, k - 1));
                a[j][k - 1] = 2;
            }

            if (k + 1 < m && a[j][k + 1] == 1) {
                q.push(make_pair(j, k + 1));
                a[j][k + 1] = 2;
            }
            q.pop();

        }

        ++timer;
    }


    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            if (a[i][j] == 1) {
                return -1;
            }
        }
    }

    return timer;

}

int main() {
    //code
    int t;
    cin >> t;
    while (t--) {
        int n, m;
        cin >> n >> m;

        vector < vector < int > > a(n, vector < int > (m, 0));
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < m; ++j) {
                cin >> a[i][j];

            }
        }

        cout << solveRotten(a, n, m) << endl;

    }

    return 0;

}