text: # 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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