Today I have done leetcode's February LeetCoding Challenge with cpp.

February LeetCoding Challenge 15

Description

The K Weakest Rows in a Matrix

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.

Solution

class Solution {
public:
    vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
        int rows = mat.size();
        int cols = mat.front().size();
        vector<int> rowCount(rows);
        for(int i = 0; i < rows; ++i) {
            int cnt = 0;
            for(auto i : mat[i]) cnt += i;
            rowCount[i] = cnt;
        }
        vector<int> answer(rows);
        for(int i = 0; i < rows; ++i) answer[i] = i;
        
        stable_sort(answer.begin(), answer.end(), [&](int a, int b){
            return rowCount[a] < rowCount[b];
        });
        answer.resize(k);
        return move(answer);
    }
};