# 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

*, if the number of soldiers in row*

**j***is less than the number of soldiers in row*

**i***, or they have the same number of soldiers but*

**j***is less than*

**i***. Soldiers are*

**j****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))
```