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+# 2530. Maximal Score After Applying K Operations
+
+This repository contains a solution to the **2530. Maximal Score After Applying K Operations** problem. The goal of the problem is to find the maximum sum we can obtain by repeatedly selecting the largest element from a list and modifying it by pushing back one-third of its value.
+
+## Problem Overview
+
+Given a list of integers `nums` and an integer `k`, the task is to maximize the sum by performing `k` operations. Each operation consists of the following steps:
+
+1. Select the largest element from the list.
+2. Add the selected element to the total sum.
+3. Replace the selected element with one-third of its value (rounded up to the nearest integer).
+4. Repeat the process `k` times.
+
+### Example
+Suppose we have the list `nums = [10, 20, 7]` and `k = 3`. The algorithm will perform the following operations:
+
+1. Select 20 (largest element), add it to the total sum, and replace it with `ceil(20/3) = 7`.
+2. Select 10 (largest element), add it to the total sum, and replace it with `ceil(10/3) = 4`.
+3. Select 7 (largest element), add it to the total sum, and replace it with `ceil(7/3) = 3`.
+
+Thus, the maximum sum we can obtain after 3 operations is `20 + 10 + 7 = 37`.
+
+## Approach
+
+This solution uses a **max-heap** (priority queue) to efficiently extract the largest element from the list and perform the operations. Here is a step-by-step breakdown of the approach:
+
+1. **Initialize a Max-Heap:** Push all elements of the input list `nums` into a max-heap (priority queue). A max-heap allows us to always extract the largest element in `O(log n)` time.
+2. **Perform K Operations:**
+   - Extract the largest element from the heap.
+   - Add it to the total sum.
+   - Replace the extracted element with one-third of its value (rounded up).
+3. **Repeat:** Continue the above steps for `k` iterations.
+
+### Time Complexity
+
+- Inserting `n` elements into the priority queue takes `O(n log n)`.
+- Each of the `k` operations involves extracting the maximum element and inserting a new value, both of which take `O(log n)` time.
+
+Thus, the overall time complexity is `O(n log n + k log n)`.
+
+## Code Explanation
+
+```cpp
+class Solution {
+public:
+    long long maxKelements(vector<int>& nums, int k) {
+        long long int ans = 0;
+        priority_queue<int> pq;
+        for (auto& i : nums) {
+            pq.push(i);
+        }
+
+        while (k--) {
+            int maxElement = pq.top();
+            ans += maxElement;
+            pq.pop();
+            pq.push(ceil(maxElement / 3.0));
+        }
+        return ans;
+    }
+};
+```
+
+### Explanation of the Code:
+- **Input:** A vector of integers `nums` and an integer `k`.
+- **Priority Queue:** A max-heap (`priority_queue<int>`) is used to store elements in descending order.
+- **Main Loop:** For each iteration (k times):
+  - Extract the largest element from the heap (`pq.top()`).
+  - Add it to the result (`ans`).
+  - Replace the extracted element with its one-third value (rounded up) and push it back into the heap (`pq.push(ceil(maxElement / 3.0))`).
+- **Return Value:** The final result (`ans`) is returned after performing the `k` operations.
+
+## Usage
+
+To use this solution, clone the repository and compile the solution with your favorite C++ compiler. You can pass different `nums` arrays and values of `k` to test the functionality.
+
+---
diff --git a/solutions/day14/solution_cpp.cpp b/solutions/day14/solution_cpp.cpp
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+class Solution {
+public:
+    long long maxKelements(vector<int>& nums, int k) {
+        long long int ans = 0;
+        priority_queue<int> pq;
+        for (auto& i : nums) {
+            pq.push(i);
+        }
+
+        while (k--) {
+            int maxElement = pq.top();
+            ans += maxElement;
+            pq.pop();
+            pq.push(ceil(maxElement / 3.0));
+        }
+        return ans;
+    }
+};
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