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:
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]
vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) { // time: O(klogk); space: O(k)
vector<vector<int> > res;
if (nums1.empty() || nums2.empty() || k == 0) return res;
auto cmp = [](const vector<int>& a, const vector<int>& b) {
return (a[0] + a[1]) > (b[0] + b[1]);
};
priority_queue<vector<int>, vector<vector<int> >, decltype(cmp)> pq(cmp); // <num1, num2, idx in num2>
// Initialization of priority queue
for (int i = 0; i < nums1.size() && i < k; ++i) pq.push({nums1[i], nums2[0], 0});
while (k-- > 0 && !pq.empty()) {
vector<int> t = pq.top();
pq.pop();
res.push_back({t[0], t[1]});
if (t[2] == nums2.size() - 1) continue;
int idx2 = t[2] + 1;
pq.push({t[0], nums2[idx2], idx2});
}
return res;
}