Question
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
Basic idea: Use min heap to keep track on next minimum pair sum, and we only need to maintain K possible candidates in the data structure.
Some observations: For every numbers in nums1, its best partner(yields min sum) always strats from nums2[0] since arrays are all sorted; And for a specific number in nums1, its next candidate sould be [this specific number] + nums2[current_associated_index + 1], unless out of boundary;)
Code
class Solution {
public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
PriorityQueue<int[]> pq = new PriorityQueue<>((p1,p2)->p1[0]+p1[1]-p2[0]-p2[1]);
List<int[]> res = new ArrayList();
if(nums2.length == 0 || nums1.length == 0 || k == 0) return res;
for(int i=0; i<k && i<nums1.length; i++) pq.add(new int[]{nums1[i], nums2[0], 0});
while(k-- > 0 && !pq.isEmpty()) {
int[] cur = pq.poll();
res.add(new int[]{cur[0], cur[1]});
if(cur[2] == nums2.length - 1) continue;
int nextIdx = cur[2]+1;
pq.add(new int[]{cur[0], nums2[nextIdx], nextIdx});
}
return res;
}
}
Performance
O(kLog(k))