Add solution to day 24

04- April/24- Last Stone Weight/24- Last Stone Weight (Ahmed Gamal).cpp

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+// Author: Ahmed Gamal
+
+// for this problem, we need to find the last stone after all the operations
+// we can use a priority queue to solve it (max heap)
+// at first, we push all the stones into the priority queue
+// then, we try to get the two heaviest stones and break them
+// if the two stones are equal, then we don't need to push anything into the priority queue
+// otherwise, we need to push the difference between the two stones into the priority queu
+
+class Solution {
+public:
+    int lastStoneWeight(vector<int>& stones) {
+        // initialize the priority queue
+        priority_queue<int> pq;
+        for(auto& i : stones) {
+            pq.push(i);
+        }
+
+        // get the top element from the priority queue
+        auto get = [&]() -> int {
+            int x = pq.top();
+            pq.pop();
+            return x;
+        };
+
+        // try to get the two heaviest stones and break them
+        while(pq.size() > 1) {
+            int y = get();
+            int x = get();
+
+            // if the two stones are equal, then we don't need to push anything into the priority queue
+            if(y - x) {
+                pq.push(y - x);
+            }
+        }
+
+        // return the answer
+        return pq.empty() ? 0 : pq.top();
+    }
+};