0373-find-k-pairs-with-smallest-sums

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Description

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

Solution(Python)

import heapq
class Solution:
    def kSmallestPairs(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:
        m = len(nums1)
        n = len(nums2)

        minHeap = [(nums1[0]+nums2[0] , (0,0))]
        heapq.heapify(minHeap)
        visited = set()
        visited.add((0,0))
        res = []
        while k > 0:
            cur_sum, (i,j) = heapq.heappop(minHeap)
            res.append([nums1[i],nums2[j]])
            

            if i + 1 < m and (i+1,j) not in visited:
                heapq.heappush(minHeap, ( nums1[i+1]+nums2[j],(i+1,j)))
                visited.add((i+1,j))
            if j+ 1 < n and (i,j + 1) not in visited:
                heapq.heappush(minHeap,( nums1[i]+nums2[j+1],(i,j+1)))
                visited.add((i,j+1))

            if i + 1 < m and j+ 1 < n and (i+1,j+1) not in visited:
                heapq.heappush(minHeap,( nums1[i+1]+nums2[j+1],(i+1,j+1)))
                visited.add((i+1,1+j))
            k-=1

        return res