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+/*
+
+Question: 2530. Maximal Score After Applying K Operations
+
+Question Description:
+You are given a 0-indexed integer array nums and an integer k. You have a starting score of 0.
+
+In one operation:
+
+choose an index i such that 0 <= i < nums.length,
+increase your score by nums[i], and
+replace nums[i] with ceil(nums[i] / 3).
+Return the maximum possible score you can attain after applying exactly k operations.
+
+The ceiling function ceil(val) is the least integer greater than or equal to val.
+
+ 
+
+Example 1:
+
+Input: nums = [10,10,10,10,10], k = 5
+Output: 50
+Explanation: Apply the operation to each array element exactly once. The final score is 10 + 10 + 10 + 10 + 10 = 50.
+Example 2:
+
+Input: nums = [1,10,3,3,3], k = 3
+Output: 17
+Explanation: You can do the following operations:
+Operation 1: Select i = 1, so nums becomes [1,4,3,3,3]. Your score increases by 10.
+Operation 2: Select i = 1, so nums becomes [1,2,3,3,3]. Your score increases by 4.
+Operation 3: Select i = 2, so nums becomes [1,1,1,3,3]. Your score increases by 3.
+The final score is 10 + 4 + 3 = 17.
+
+*/
+
+class Solution {
+public:
+    long long maxKelements(vector<int>& nums, int k) {
+        
+
+        long long score = 0;    
+        priority_queue<int> pq;     //create a priority queue "pq"
+        for(int i=0; i < nums.size(); i++){     //add the elements to pq
+            pq.push(nums[i]);
+        }
+
+        while(k > 0){
+            int max = pq.top();
+            if(max == 1)    //this condition helps to exit early hence more efficient
+                return score + k;
+            score += max;   
+            pq.pop();
+            pq.push(ceil(max / 3.0));   //insert this element in place of max
+            k--;
+        }
+
+        return score;
+
+    }
+};
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