adding another leetcode pq problem

leetcode/p373_k_pairs_smallest_sums.cpp

@@ -0,0 +1,44 @@
+class cmp {
+public:
+    // overloading the function call operator
+    bool operator()(pair<int,pair<int,int>> p1, pair<int,pair<int,int>> p2){
+        return p1.first < p2.first;
+    }
+};
+
+class Solution {
+public:
+    vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
+        int size1 = nums1.size();
+        int size2 = nums2.size();
+        vector<vector<int> > ans;
+        priority_queue<pair<int,pair<int,int>>,vector<pair<int,pair<int,int>>>,cmp> pq;
+        for(int i=0; i<size1; i++){
+            for(int j=0; j<size2; j++){
+                int sum=nums1[i]+nums2[j];
+                
+                if(pq.size()<k){
+                    pq.push( {sum , {nums1[i] , nums2[j]} } );
+                }
+                // sum is smaller than the sum stored at top(), which is the max in the heap, then let that max go;
+                else if(sum < pq.top().first){
+                    pq.pop();
+                    pq.push( {sum , {nums1[i] , nums2[j]} } );
+                }else{
+                    // else means, we have k elements in the heap, and sum is larger than the max; then all remaining indices (in this current row) would just generate larger sums. 
+                    // so this break is not breaking out of the entire nested loops, but rather, just break out of the inner loop, because we now know matrix[i][j] is already excluded, then of course, matrix[i][j+1], matrix[i][j+2], etc., should all be excluded.
+                    break;
+                }
+            }
+        }
+        
+        while(!pq.empty()){
+            // this is a max heap, so every time we get is the largest; therefore data in ans is sorted (based on the sum) in descending order
+            ans.push_back({pq.top().second.first , pq.top().second.second});
+            pq.pop();
+        }
+        // reverse it so we get it in ascending order.
+        reverse(ans.begin(), ans.end());
+        return ans;
+    }
+};