Find K Pairs with Smallest Sums

#leetcode #dsa #theabbie

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:

import heapq

class Solution:
    def kSmallestPairs(self, nums1, nums2, k: int):
        heap = []
        a = min(len(nums1), k)
        b = min(len(nums2), k)
        for i in range(a):
            for j in range(b):
                heapq.heappush(heap, (-nums1[i]-nums2[j], (nums1[i], nums2[j])))
                if len(heap) > k:
                    heapq.heappop(heap)
        ksmallest = []
        for i in range(k):
            if len(heap) > 0:
                ksmallest.insert(0, list(heapq.heappop(heap)[1]))
        return ksmallest