# Find the Kth Smallest Sum of a Matrix With Sorted Rows

You are given an m * n matrix, mat, and an integer k, which has its rows sorted in non-decreasing order.

You are allowed to choose exactly 1 element from each row to form an array. Return the Kth smallest array sum among all possible arrays.

Example 1:

```Input: mat = [[1,3,11],[2,4,6]], k = 5
Output: 7
Explanation: Choosing one element from each row, the first k smallest sum are:
[1,2], [1,4], [3,2], [3,4], [1,6]. Where the 5th sum is 7.  ```
Example 2:

```Input: mat = [[1,3,11],[2,4,6]], k = 9
Output: 17
```
Example 3:

```Input: mat = [[1,10,10],[1,4,5],[2,3,6]], k = 7
Output: 9
Explanation: Choosing one element from each row, the first k smallest sum are:
[1,1,2], [1,1,3], [1,4,2], [1,4,3], [1,1,6], [1,5,2], [1,5,3]. Where the 7th sum is 9.
```
Example 4:

```Input: mat = [[1,1,10],[2,2,9]], k = 7
Output: 12
```

Constraints:

• m == mat.length
• n == mat.length[i]
• 1 <= m, n <= 40
• 1 <= k <= min(200, n ^ m)
• 1 <= mat[i][j] <= 5000
• mat[i] is a non decreasing array.

Solution:

```class Solution {
public int kthSmallest(int[][] mat, int k) {
int m = mat.length, n = mat.length;
PriorityQueue<int[]> pq = new PriorityQueue<int[]>((a, b) -> {
int sumA = 0, sumB = 0;
for (int i = 0; i < m; i ++) {
sumA += mat[i][a[i]];
sumB += mat[i][b[i]];
}
return Integer.compare(sumA, sumB);
});
Set<List<Integer>> set = new HashSet();
pq.offer(new int[m]);
set.add(Arrays.stream( new int[m] ).boxed().collect( Collectors.toList() ));
int res = 0;
while (k > 1) {
int[] curr = pq.poll();
for (int i = 0; i < m; i ++) {
int[] next = curr.clone();
if (curr[i] + 1 < n) {
next[i] ++;
}
if (set.add(Arrays.stream( next ).boxed().collect( Collectors.toList() ))) {
pq.offer(next);
}
}
k --;
}
int[] curr = pq.poll();
for (int i = 0; i < m; i ++) {
res += mat[i][curr[i]];
}
return res;
}
}```