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The K Weakest Rows in a Matrix

Updated: Mar 25, 2021


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.



class Solution {
    public int[] kWeakestRows(int[][] mat, int k) {
        PriorityQueue<int[]> pq = new PriorityQueue<>((a,b)-> a[1] ==b[1]?b[0]-a[0]:b[1]-a[1]);
        
        for(int i=0;i<mat.length;i++)
        {
            int j=0;
            while(j<mat[0].length && mat[i][j]==1) j++;
            pq.add(new int[]{i,j});
            if(pq.size()>k)
            {
                pq.poll();
            }
        }
        int[] result = new int[k];
        while(k>0)
        {
            result[--k] = pq.poll()[0];
        }
        return result;
    }
}
//O(r*(c+logk))


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