1341

1341. The K Weakest Rows in a Matrix (Easy)

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.

Related Topics:
Array, Binary Search

Solution 1.

// OJ: https://leetcode.com/problems/the-k-weakest-rows-in-a-matrix/
// Author: github.com/lzl124631x
// Time: O(MN + MlogM)
// Space: O(M)
class Solution {
public:
    vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
        vector<pair<int, int>> v;
        for (int i = 0; i < mat.size(); ++i) {
            int cnt = 0;
            for (int j = 0; j < mat[i].size() && mat[i][j]; ++j) ++cnt;
            v.emplace_back(cnt, i);
        }
        sort(v.begin(), v.end());
        vector<int> ans;
        for (int i = 0; i < k; ++i) ans.push_back(v[i].second);
        return ans;
    }
};