## 373. Find K Pairs with Smallest Sums

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 = , k = 3

Output: [1,3],[2,3]
Explanation: All possible pairs are returned from the sequence: [1,3],[2,3]
``````

Thoughts:

1. Min Sum: min element from num1 and nums2. Smallest is definitely nums1 + num2
2. Selecting the next candidate needs extra comparison: nums1[i] + nums2[j + 1] vs nums1[i + 1] + nums2[j], and then it also needs to compare the candidate.
3. Thus, use priority queue to maintain the order of the sums. In order to add duplicates, first offer k sums with nums1[i] + nums2 for i = 0,...k. Code: T:O(klogk); S: O(k)

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