You are given two integer arraysnums1andnums2sorted in ascending order and an integerk.

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]

class Solution {
    public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
        List<int[]> ans = new ArrayList<>();
        if (nums1 == null || nums2 == null || nums1.length == 0 || nums2.length == 0 || k == 0) return ans;

        PriorityQueue<int[]> pq = new PriorityQueue<int[]>((a, b) -> (a[0] + a[1] - b[0] - b[1]));

        for (int i = 0; i < nums1.length; i++){
            pq.offer(new int[]{nums1[i], nums2[0], 0});
        }

        while (k-- > 0 && !pq.isEmpty()){
            int[] cur = pq.poll();
            ans.add(new int[]{cur[0], cur[1]});
            if (cur[2] == nums2.length - 1) continue;
            pq.offer(new int[]{cur[0], nums2[cur[2] + 1], cur[2] + 1});
        }

        return ans;
    }
}