# Find K Pairs with Smallest Sums LeetCode Solution

Here, We see Find K Pairs with Smallest Sums LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

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]

## Find K Pairs with Smallest Sums LeetCode SolutionC++

``````class Solution {
public:
vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
vector<vector<int>> resV;
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
for(int x : nums1) {
pq.push({x + nums2[0], 0});
}
while(k-- && !pq.empty()) {
int sum = pq.top().first;
int pos = pq.top().second;
resV.push_back({sum - nums2[pos], nums2[pos]});
pq.pop();
if(pos + 1 < nums2.size()) {
pq.push({sum - nums2[pos] + nums2[pos + 1], pos + 1});
}
}
return resV;
}
};```Code language: PHP (php)```

## Find K Pairs with Smallest Sums Leetcode SolutionJava

``````class Solution {
public List<List<Integer>> kSmallestPairs(int[] nums1, int[] nums2, int k) {
List<List<Integer>> resV = new ArrayList<>();
PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> a[0] - b[0]);
for (int x : nums1) {
pq.offer(new int[]{x + nums2[0], 0});
}
while (k > 0 && !pq.isEmpty()) {
int[] pair = pq.poll();
int sum = pair[0];
int pos = pair[1];
List<Integer> currentPair = new ArrayList<>();
if (pos + 1 < nums2.length) {
pq.offer(new int[]{sum - nums2[pos] + nums2[pos + 1], pos + 1});
}
k--;
}
return resV;
}
}```Code language: PHP (php)```

## Find K Pairs with Smallest Sums SolutionJavaScript

``````var kSmallestPairs = function(nums1, nums2, k) {
const minHeap = new MinPriorityQueue({ priority: x => x[0] });
for (let i = 0; i < nums1.length; i++) {
const num1 = nums1[i];
const num2 = nums2[0];
minHeap.enqueue([num1 + num2, i, 0]);
}
const n = nums2.length;
const res = [];
while (k > 0 && !minHeap.isEmpty()) {
const [sum, idx1, idx2] = minHeap.dequeue().element;
res.push([nums1[idx1], nums2[idx2]]);
if (res.length === k) return res;
if (idx2 < n - 1) {
minHeap.enqueue([nums1[idx1] + nums2[idx2 + 1], idx1, idx2 + 1]);
}
}
return res;
};```Code language: JavaScript (javascript)```

## Find K Pairs with Smallest Sums SolutionPython

``````class Solution(object):
def kSmallestPairs(self, nums1, nums2, k):
resV = []
pq = []
for x in nums1:
heapq.heappush(pq, [x + nums2[0], 0])
while k > 0 and pq:
pair = heapq.heappop(pq)
s, pos = pair[0], pair[1]
resV.append([s - nums2[pos], nums2[pos]])
if pos + 1 < len(nums2):
heapq.heappush(pq, [s - nums2[pos] + nums2[pos + 1], pos + 1])
k -= 1
return resV```Code language: HTML, XML (xml)```
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