# Question

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:

Given nums1 = [1,7,11], nums2 = [2,4,6],  k = 3

Return: [1,2],[1,4],[1,6]

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:

Given nums1 = [1,1,2], nums2 = [1,2,3],  k = 2

Return: [1,1],[1,1]

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:

Given nums1 = [1,2], nums2 = [3],  k = 3

Return: [1,3],[2,3]

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


# Solution

We can use a heap to store k candidate pairs. For every numbers in nums1, its best partner (yields min sum) always starts from nums2[0] since arrays are all sorted; And for a specific number in nums1, its next candidate should be [this specific number] + nums2[current_partner_index + 1], unless out of boundary for nums2.

class Solution {
public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
k = Math.min(nums1.length * nums2.length, k);
List<int[]> res = new ArrayList<>(k);
PriorityQueue<int[]> q = new PriorityQueue<>((a, b) -> (a[0] + a[1]) - (b[0] + b[1]));
int l = 0, r = 0;
for(int i = 0; i < Math.min(k, nums1.length); i++)
q.offer(new int[]{nums1[i], nums2[0], 0});
for(int i = 0; i < k; i++) {
int[] arr = q.poll();
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