Create kth_smallest_element_in_sorted_matrix.cpp

Author: avidLearnerInProgress

kth_smallest_element_in_sorted_matrix.cpp

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+class Solution {
+public:
+    //TC: O(r*c*log k) | SC: O(k)
+    //This approach doesnt use the advantage of sorted property of rows and columns
+//     int kthSmallest(vector<vector<int>>& matrix, int k) {
+//         if(matrix.size() == 0 or k == 0) return -1;
+    
+//         priority_queue<int> maxHeap;
+        
+//         for(int r = 0; r < matrix.size(); r++) {
+//             for(int c = 0; c < matrix[r].size(); c++) {
+//                 maxHeap.emplace(matrix[r][c]);
+//                 if(maxHeap.size() > k) maxHeap.pop();
+//             }
+//         }
+//         return maxHeap.top();
+//     }
+    
+    
+    //since all the rows and cols are sorted, we can observe the problems as finding kth smallest element from sorted rows using min heap
+    //optimization based on this post - https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/1322101/C%2B%2BJavaPython-MaxHeap-MinHeap-Binary-Search-Picture-Explain-Clean-and-Concise
+    
+    //TC: O(k * log k) | SC: O(k)
+    int kthSmallest(vector<vector<int>>& matrix, int k) {
+        if(matrix.size() == 0 or k == 0) return -1;
+        int m = matrix.size(), n = matrix[0].size();
+        
+        priority_queue<vector<int>, vector<vector<int>>, greater<> > minHeap;
+        for(int r = 0; r < min(m, k); ++r) { //min() because we always know that the kth smallest element will lie in min(total rows, kth row)
+            minHeap.push({matrix[r][0], r, 0});    
+        }
+        
+        int kthSmallest = -1;
+        for(int i = 1; i <= k; ++i) {
+            auto ele = minHeap.top();
+            kthSmallest = ele[0];
+            int r = ele[1];
+            int c = ele[2];
+            
+            minHeap.pop();
+            if(c + 1 < n) {
+                minHeap.push({matrix[r][c+1], r, c+1});
+            }
+        }
+        return kthSmallest;
+    }
+    
+};