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))
```

## ComentÃ¡rios