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373. Find K Pairs with Smallest Sums

Difficulty Topics

Description

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.

Return 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]

 

Constraints:

  • 1 <= nums1.length, nums2.length <= 105
  • -109 <= nums1[i], nums2[i] <= 109
  • nums1 and nums2 both are sorted in ascending order.
  • 1 <= k <= 104

Solution

find-k-pairs-with-smallest-sums.py
class Solution:
    def kSmallestPairs(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:
        n1, n2 = len(nums1), len(nums2)
        pq = []
        res = []

        for i in range(min(n1, k)):
            heappush(pq, ((nums1[i] + nums2[0]), i, 0))

        for _ in range(k):
            if not pq: break

            _, i, j = heappop(pq)
            res.append([nums1[i], nums2[j]])
            if j == n2 - 1: continue
            heappush(pq, (nums1[i] + nums2[j + 1], i, j + 1))

        return res