Coding

1337. The K Weakest Rows in a Matrix

2023-09-18 2 min read
#Array#Binary Search#Sorting#Heap (Priority Queue)#Matrix
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題目敘述

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].

解題思路

用 sorting 去完成。

Solution

import java.util.Arrays;

class Solution {
    public int[] kWeakestRows(int[][] mat, int k) {
        int m = mat.length;
        int[][] rowAdd = new int[m][2];

        for(int i = 0; i < m; i++){
            int add = 0;
            for(int num : mat[i]){
                add += num;
            }
            rowAdd[i][0] = add;
            rowAdd[i][1] = i;
        }
        
        Arrays.sort(rowAdd, (a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);

        int[] ans = new int[k];
        for(int i = 0; i < k; i++){
            ans[i] = rowAdd[i][1];
        }

        return ans;
    }
}
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