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 <= 10^5
- -109 <= nums1[i], nums2[i] <= 10^9
- nums1 and nums2 both are sorted in ascending order.
- 1 <= k <= 10^4

class Solution {
public:
    vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
        vector<vector<int>> result;

        priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > pq;

        for (int i = 0; i < nums1.size(); ++i)
            pq.push({nums1[i] + nums2[0], 0}); // Sum and index of the second element

        while (k-- && !pq.empty()) {
            int sum = pq.top().first;
            int indx = pq.top().second;
            result.push_back({sum — nums2[indx], nums2[indx]});
            pq.pop();

            if (indx + 1 < nums2.size())
                pq.push({sum — nums2[indx] + nums2[indx + 1], indx + 1});
        }
        return result;
    }
};