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Abhishek Chaudhary

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# 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.

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
``````