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:
public class Solution {
public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
List<int[]> ans = new ArrayList<>();
if(nums1.length==0 || nums2.length==0 || k==0) return ans;
PriorityQueue<int[]> pq = new PriorityQueue<>(k, (a,b)->((nums1[a[0]]+nums2[a[1]])-(nums1[b[0]]+nums2[b[1]])));
for(int i=0; i<nums1.length; i++) pq.offer(new int[]{i,0});
while(k-->0 && pq.size()!=0){
int[] pair = pq.poll();
ans.add(new int[]{nums1[pair[0]], nums2[pair[1]]});
if(pair[1]<nums2.length-1) pq.offer(new int[]{pair[0],pair[1]+1});
}
return ans;
}
}