1337. The K Weakest Rows in a Matrix.cpp

Given a m * n matrix mat of ones (representing soldiers) and zeros (representing civilians), return the indexes of the k weakest rows in the matrix ordered from the weakest to the strongest.

A row i is weaker than row j, if the number of soldiers in row i is less than the number of soldiers in row j, or they have the same number of soldiers but i is less than j. Soldiers are always stand in the frontier of a row, that is, always ones may appear first and then zeros.



Example 1:

Input: mat =
[[1,1,0,0,0],
 [1,1,1,1,0],
 [1,0,0,0,0],
 [1,1,0,0,0],
 [1,1,1,1,1]],
k = 3
Output: [2,0,3]
Explanation:
The number of soldiers for each row is:
row 0 -> 2
row 1 -> 4
row 2 -> 1
row 3 -> 2
row 4 -> 5
Rows ordered from the weakest to the strongest are [2,0,3,1,4]
Example 2:

Input: mat =
[[1,0,0,0],
 [1,1,1,1],
 [1,0,0,0],
 [1,0,0,0]],
k = 2
Output: [0,2]
Explanation:
The number of soldiers for each row is:
row 0 -> 1
row 1 -> 4
row 2 -> 1
row 3 -> 1
Rows ordered from the weakest to the strongest are [0,2,3,1]


Constraints:

m == mat.length
n == mat[i].length
2 <= n, m <= 100
1 <= k <= m
matrix[i][j] is either 0 or 1.






class Solution {
public:

    static bool comp(vector<int> a, vector<int> b){
        if(a[0]==b[0]) return a[1]<b[1];
        return a[0]<b[0];
    }

    vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
        vector<int> res;

        int m=mat.size();
        int n=mat[0].size();

        vector<vector<int> > rows(m, vector<int> (2,0));

        for(int i=0;i<m;i++){
            int ones=0;
            for(int j=0;j<n;j++){
                if(mat[i][j]==1) ones++;
                else break;
            }
            rows[i][0]=ones;
            rows[i][1]=i;
        }

        sort(rows.begin(), rows.end(), comp);

        for(int i=0;i<k;i++)
            res.push_back(rows[i][1]);

        return res;
    }
};