1337. The K Weakest Rows in a Matrix

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 the k weakest rows in the matrix ordered from weakest to strongest.

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 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:

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].

Constraint:

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

Solution (Java)

class Solution {
    public int[] kWeakestRows(int[][] mat, int k) {
        int[] result = new int[mat.length];
        for (int i = 0; i < mat.length; i++) {
            int index = binarySearch(mat, i, mat[i].length - 1);
            result[i] = index;
        }
        int minValue = 101;
        int[] resultK = new int[k];
        int index = -1;
        for (int i = 0; i < k; i++) {
            for (int j = 0; j < result.length; j++) {
                if (result[j] < minValue) {
                    minValue = result[j];
                    index = j;
                }
            }
            result[index] = 110;
            resultK[i] = index;
            index = -1;
            minValue = 101;
        }
        return resultK;
    }

    private int binarySearch(int[][] mat, int row, int end) {
        int start = 0;
        while (start <= end) {
            int mid = start + (end - start) / 2;
            if (mat[row][mid] == 1) {
                start = mid + 1;
            } else if (mat[row][mid] == 0) {
                end = mid - 1;
            }
        }
        return start;
    }
}

Solution (Javascript)

/**
 * @param {number[][]} mat
 * @param {number} k
 * @return {number[]}
 */
var kWeakestRows = function(mat, k) {
    let y = mat.length, x = mat[0].length,
        vis = new Uint8Array(y), ans = []
    for (let j = 0; j <= x; j++)
        for (let i = 0; i < y; i++) {
            if (!vis[i] && !mat[i][j]) ans.push(i), vis[i]++
            if (ans.length === k) return ans
        }
};