Find K Pairs with Smallest Sums


#1

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]

#2

We here use a priority queue to maintain a min heap of sums.

class Solution {
    public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
       PriorityQueue<int[]> pq = new PriorityQueue<>(new Comparator<int[]>(){
            public int compare(int[] a , int [] b){
                if(a[0] + a[1] == b[0] + b[1]){
                    return a[0] - b[0];
                }
                return a[0] + a[1] - (b[0] + b[1]);
            }
        });
        List<int[]> result = new ArrayList<>();
        if(nums1.length==0 || nums2.length==0 || k==0) return result;
        for(int i=0;i<nums1.length && i<k; i++) pq.offer(new int[]{nums1[i], nums2[0], 0});
         int[] cur = new int[3];
        while(k-- > 0 && !pq.isEmpty()){
            cur = pq.poll();
            result.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 result;
    }
}