## Find K Pairs with Smallest Sums

• Time:O(k\log k)
• Space:O(k)

## C++

``````struct T {
int i;
int j;
int sum;  // nums1[i] + nums2[j];
T(int i, int j, int sum) : i(i), j(j), sum(sum) {}
};

class Solution {
public:
vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2,
int k) {
vector<vector<int>> ans;
auto compare = [&](const T& a, const T& b) { return a.sum > b.sum; };
priority_queue<T, vector<T>, decltype(compare)> minHeap(compare);

for (int i = 0; i < k && i < nums1.size(); ++i)
minHeap.emplace(i, 0, nums1[i] + nums2[0]);

while (!minHeap.empty() && ans.size() < k) {
const auto [i, j, _] = minHeap.top();
minHeap.pop();
ans.push_back({nums1[i], nums2[j]});
if (j + 1 < nums2.size())
minHeap.emplace(i, j + 1, nums1[i] + nums2[j + 1]);
}

return ans;
}
};
``````

## JAVA

``````class T {
public int i;
public int j;
public int sum; // nums1[i] + nums2[j]
public T(int i, int j, int sum) {
this.i = i;
this.j = j;
this.sum = sum;
}
}

class Solution {
public List<List<Integer>> kSmallestPairs(int[] nums1, int[] nums2, int k) {
List<List<Integer>> ans = new ArrayList<>();
Queue<T> minHeap = new PriorityQueue<>((a, b) -> a.sum - b.sum);

for (int i = 0; i < k && i < nums1.length; ++i)
minHeap.offer(new T(i, 0, nums1[i] + nums2[0]));

while (!minHeap.isEmpty() && ans.size() < k) {
final int i = minHeap.peek().i;
final int j = minHeap.poll().j;