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373. Find K Pairs With Smallest Sums

Array Heap (Priority Queue)

Problem - Find K Pairs With Smallest Sums

Medium

You are given two integer arrays nums1 and nums2 sorted in non-decreasing 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]

 

Constraints:

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

Solutions

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class Solution:
    def kSmallestPairs(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:
        queue = [[val + nums2[0], i, 0] for i, val in enumerate(nums1[:k])]
        result = []
        heapify(queue)

        while queue and k > 0:
            _, i, j = heappop(queue)
            result.append([nums1[i], nums2[j]])
            k -= 1

            if j + 1 < len(nums2):
                heappush(queue, [nums1[i] + nums2[j + 1], i, j + 1])

        return result

Submission Stats:

  • Runtime: 91 ms (67.91%)
  • Memory: 34.7 MB (70.88%)