1337. The K Weakest Rows in a Matrix

Description

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.

Constraints

• m == mat.length

• n == mat[i].length

• 2 <= n, m <= 100

• 1 <= k <= m

• matrix[i][j] is either 0 or 1.

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int[] kWeakestRows(int[][] mat, int k) {
        Map<Integer, Integer> map = new HashMap();
        int m = mat.length;
        
        for(int i = 0; i < m; i++) {
            map.put(i, getLastIndex(mat[i], 0, mat[i].length-1)+1);
        }
        
        PriorityQueue<Integer> pq = new PriorityQueue<>(
            (Integer val1, Integer val2) -> {
                int diff = map.get(val1)-map.get(val2);
                return (diff == 0)? val1-val2: diff;
            }
        );
        
        for(int i = 0; i < m; i++) {
            pq.add(i);
        }
        
        int[] result = new int[k];
        for(int i = 0; i < k; i++) {
            result[i] = pq.poll();
        }
        
        return result;
    }
    
    private int getLastIndex(int[] arr, int left, int right) {
        int index = -1;
        while(left <= right) {
            int mid = (left + right) / 2;
            if(arr[mid] == 1) {
                index = Math.max(index, mid);
                left = mid+1;
            } else {
                right = mid-1;
            }
        }
        return index;
    }
}

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