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 = [3], 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[0] + num2[0]
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[0] 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[0] + a[1] - b[0] - b[1]);
        for(int i = 0; i < nums1.length && i < k; i++) pq.offer(new int []{nums1[i], nums2[0], 0}); // end of nums1
        while(k-- > 0 && !pq.isEmpty()){
            int [] cur = pq.poll();
            res.add(new int[]{cur[0], cur[1]});
            if(cur[2] == nums2.length - 1) continue; // end of nums2
            pq.offer(new int[] {cur[0], nums2[cur[2] + 1], cur[2] + 1});
        }

        return res;
    }
}