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
We can use a heap to store k
candidate pairs. For every numbers in nums1
, its best partner (yields min sum) always starts from nums2[0]
since arrays are all sorted; And for a specific number in nums1
, its next candidate should be [this specific number] + nums2[current_partner_index + 1]
, unless out of boundary for nums2
.
class Solution {
public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
k = Math.min(nums1.length * nums2.length, k);
List<int[]> res = new ArrayList<>(k);
PriorityQueue<int[]> q = new PriorityQueue<>((a, b) -> (a[0] + a[1]) - (b[0] + b[1]));
int l = 0, r = 0;
for(int i = 0; i < Math.min(k, nums1.length); i++)
q.offer(new int[]{nums1[i], nums2[0], 0});
for(int i = 0; i < k; i++) {
int[] arr = q.poll();
res.add(new int[]{arr[0], arr[1]});
if(arr[2] < nums2.length - 1)
q.offer(new int[]{arr[0], nums2[++arr[2]], arr[2]});
}
return res;
}
}