373. Find K Pairs with Smallest Sums

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: Heap

import heapq


class Solution(object):
    def kSmallestPairs(self, nums1, nums2, k):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :type k: int
        :rtype: List[List[int]]
        """
        q = []
        for i, num1 in enumerate(nums1[:1]):
            for j, num2 in enumerate(nums2[:k]):
                heapq.heappush(q, (num1 + num2, i, j))
        pairs = []
        while len(pairs) < k and q:
            _, i, j = heapq.heappop(q)
            num1 = nums1[i]
            num2 = nums2[j]
            pairs.append([num1, num2])
            if i < len(nums1) - 1:
                i += 1
                heapq.heappush(q, (nums1[i] + num2, i, j))
        return pairs

Lessons:

• Push the first number in nums1 and all numbers in nums2 to heap.
• After each pop, add the next number nums1 and the current number in nums2 to heap.
• O(k) space, O(klogk) time.

Related:

• 378. Kth Smallest Element in a Sorted Matrix