LeetCode-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) ...(u_k,v_k) 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]

Credits:
Special thanks to @elmirap and @StefanPochmann for adding this problem and creating all test cases.

 

Solution:

public class Solution {
    public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
        List<int[]> res = new ArrayList<int[]>();
    
    // corner case
    if (nums1.length == 0 || nums2.length == 0) return res;
    
    // priority queue comparing the pair sum
    PriorityQueue<int[]> pairs = new PriorityQueue<int[]>(
        new Comparator<int[]>() {
            public int compare(int[] v1, int[] v2){
                return (v1[0] - v2[0]);
            }
        });
    
    // for each num in nums1, increase the position in nums2 from 0
    for (int i = 0; i < nums1.length; i++) {
        int[] v = {nums1[i] + nums2[0], i, 0};
        pairs.add(v);
    }
    
    while (!pairs.isEmpty() && k > 0) {
        int[] v = pairs.poll();
        // move to the next position in nums2
        if (v[2] < nums2.length - 1)    {
            int[] nv = {nums1[v[1]] + nums2[v[2] + 1], v[1], v[2] + 1};
            pairs.add(nv);
        }
        int[] pair = {nums1[v[1]], nums2[v[2]]};
        res.add(pair);
        
        k--;
    }
    
    return res;
    }
}