# Find K Pairs with Smallest Sums

#1

You are given two integer arrays nums1 and nums2 sorted in ascending order and an integer k.

Define a pair (u,v) which consists of one element from the first array and one element from the second array.

Find the k pairs (u1,v1),(u2,v2) ...(uk,vk) with the smallest sums.

Example 1:

```Input: nums1 = [1,7,11], nums2 = [2,4,6], k = 3
Output: [[1,2],[1,4],[1,6]]
Explanation: The first 3 pairs are returned from the sequence:
[1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6]```

Example 2:

```Input: nums1 = [1,1,2], nums2 = [1,2,3], k = 2
Output: [1,1],[1,1]
Explanation: The first 2 pairs are returned from the sequence:
[1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]```

Example 3:

```Input: nums1 = [1,2], nums2 = [3], k = 3
Output: [1,3],[2,3]
Explanation: All possible pairs are returned from the sequence: [1,3],[2,3]
```

#2

We here use a priority queue to maintain a min heap of sums.

``````class Solution {
public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
PriorityQueue<int[]> pq = new PriorityQueue<>(new Comparator<int[]>(){
public int compare(int[] a , int [] b){
if(a[0] + a[1] == b[0] + b[1]){
return a[0] - b[0];
}
return a[0] + a[1] - (b[0] + b[1]);
}
});
List<int[]> result = new ArrayList<>();
if(nums1.length==0 || nums2.length==0 || k==0) return result;
for(int i=0;i<nums1.length && i<k; i++) pq.offer(new int[]{nums1[i], nums2[0], 0});
int[] cur = new int[3];
while(k-- > 0 && !pq.isEmpty()){
cur = pq.poll();