Problem

You are given an m x n binary matrix mat of 1’s (representing soldiers) and 0’s (representing civilians). The soldiers are positioned in front of the civilians. That is, all the 1’s will appear to the left of all the 0’s in each row.

A row i is weaker than a row j if one of the following is true:

  • The number of soldiers in row i is less than the number of soldiers in row j.
  • Both rows have the same number of soldiers and i < j.

Return the indices of thek weakest rows in the matrix ordered from weakest to strongest.

Examples

Example 1

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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 in each row is: 
- Row 0: 2 
- Row 1: 4 
- Row 2: 1 
- Row 3: 2 
- Row 4: 5 
The rows ordered from weakest to strongest are [2,0,3,1,4].

Example 2

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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 in each row is: 
- Row 0: 1 
- Row 1: 4 
- Row 2: 1 
- Row 3: 1 
The rows ordered from weakest to 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

Method 1 - Count and Sort

Count the number of soldiers in each row, then sort rows by (soldier count, row index) and return the first k indices.

Code

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#include <vector>
#include <algorithm>
using namespace std;
vector<int> kWeakestRows(vector<vector<int>>& mat, int k) {
    vector<pair<int,int>> v;
    for (int i = 0; i < mat.size(); ++i) {
        int cnt = count(mat[i].begin(), mat[i].end(), 1);
        v.emplace_back(cnt, i);
    }
    sort(v.begin(), v.end());
    vector<int> res;
    for (int i = 0; i < k; ++i) res.push_back(v[i].second);
    return res;
}
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import java.util.*;
class Solution {
    public int[] kWeakestRows(int[][] mat, int k) {
        int m = mat.length, n = mat[0].length;
        int[][] arr = new int[m][2];
        for (int i = 0; i < m; ++i) {
            int cnt = 0;
            for (int j = 0; j < n; ++j) cnt += mat[i][j];
            arr[i][0] = cnt;
            arr[i][1] = i;
        }
        Arrays.sort(arr, (a, b) -> a[0] != b[0] ? a[0] - b[0] : a[1] - b[1]);
        int[] res = new int[k];
        for (int i = 0; i < k; ++i) res[i] = arr[i][1];
        return res;
    }
}
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from typing import List
def kWeakestRows(mat: List[List[int]], k: int) -> List[int]:
    v = [(sum(row), i) for i, row in enumerate(mat)]
    v.sort()
    return [i for _, i in v[:k]]
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pub fn k_weakest_rows(mat: Vec<Vec<i32>>, k: i32) -> Vec<i32> {
    let mut v: Vec<(i32, i32)> = mat.iter().enumerate().map(|(i, row)| (row.iter().sum(), i as i32)).collect();
    v.sort();
    v.iter().take(k as usize).map(|&(_, i)| i).collect()
}
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function kWeakestRows(mat: number[][], k: number): number[] {
    const v = mat.map((row, i) => [row.reduce((a, b) => a + b, 0), i]);
    v.sort((a, b) => a[0] - b[0] || a[1] - b[1]);
    return v.slice(0, k).map(x => x[1]);
}

Complexity

  • ⏰ Time complexity: O(mn + m log m)
  • 🧺 Space complexity: O(m)