Add Solution to LC problem 3040

dsa-solutions/lc-solutions/3000-3099/3040 - Maximum Number of Operations With the Same Score II.md

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+---
+id: maximum-number-of-operations-with-the-same-score-ii
+title: Maximum Number of Operations With the Same Score II
+sidebar_label: 3040 - Maximum Number of Operations With the Same Score II
+tags: [Array, Greedy]
+description: Find the maximum number of operations with the same score on an array of integers using specific deletion operations.
+---
+
+## Problem Statement
+
+### Problem Description
+
+Given an array of integers called `nums`, you can perform any of the following operations while `nums` contains at least 2 elements:
+
+1. Choose the first two elements of `nums` and delete them.
+2. Choose the last two elements of `nums` and delete them.
+3. Choose the first and the last elements of `nums` and delete them.
+
+The score of the operation is the sum of the deleted elements.
+
+Your task is to find the maximum number of operations that can be performed, such that all operations have the same score.
+
+Return the maximum number of operations possible that satisfy the condition mentioned above.
+
+### Example
+
+**Example 1:**
+```
+Input: nums = [3,2,1,2,3,4]
+Output: 3
+```
+**Explanation:** We perform the following operations:
+
+- Delete the first two elements, with score 3 + 2 = 5, nums = [1,2,3,4].
+- Delete the first and the last elements, with score 1 + 4 = 5, nums = [2,3].
+- Delete the first and the last elements, with score 2 + 3 = 5, nums = [].
+- We are unable to perform any more operations as nums is empty.
+
+
+### Constraints
+
+- 2 ≤ `nums.length` ≤ 2000
+- 1 ≤ `nums[i]` ≤ 1000
+
+## Solution
+
+### Intuition
+
+To solve this problem, we need to identify how many operations can be performed with the same score. We can use the following approach:
+
+1. **Calculate Possible Scores:** Compute possible scores for each operation type (first two elements, last two elements, first and last elements).
+2. **Count Operations for Each Score:** Use a dictionary to count the number of times each score can be achieved by applying each operation type.
+3. **Find the Maximum Valid Score:** Determine the maximum count of operations where all operations yield the same score.
+
+### Time and Space Complexity
+
+- **Time Complexity:** The time complexity is $O(n)$, where $n$ is the length of `nums`. This includes calculating possible scores and counting their frequencies.
+
+- **Space Complexity:** The space complexity is $O(n)$ due to the use of a dictionary to store score counts.
+
+### Code
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int maxOperations(vector<int>& nums) {
+        int n = nums.size();
+        unordered_map<int, int> scoreCount;
+
+        int maxOps = 0;
+        for (int i = 0; i < n / 2; ++i) {
+            int score1 = nums[i] + nums[i + 1];
+            int score2 = nums[n - 1 - i] + nums[n - 2 - i];
+            int score3 = nums[i] + nums[n - 1 - i];
+
+            scoreCount[score1]++;
+            scoreCount[score2]++;
+            scoreCount[score3]++;
+
+            maxOps = max(maxOps, max({scoreCount[score1], scoreCount[score2], scoreCount[score3]}));
+        }
+
+        return maxOps;
+    }
+};
+```
+
+#### Java
+```java
+import java.util.HashMap;
+import java.util.Map;
+
+class Solution {
+    public int maxOperations(int[] nums) {
+        int n = nums.length;
+        Map<Integer, Integer> scoreCount = new HashMap<>();
+
+        int maxOps = 0;
+        for (int i = 0; i < n / 2; ++i) {
+            int score1 = nums[i] + nums[i + 1];
+            int score2 = nums[n - 1 - i] + nums[n - 2 - i];
+            int score3 = nums[i] + nums[n - 1 - i];
+
+            scoreCount.put(score1, scoreCount.getOrDefault(score1, 0) + 1);
+            scoreCount.put(score2, scoreCount.getOrDefault(score2, 0) + 1);
+            scoreCount.put(score3, scoreCount.getOrDefault(score3, 0) + 1);
+
+            maxOps = Math.max(maxOps, Math.max(scoreCount.get(score1), Math.max(scoreCount.get(score2), scoreCount.get(score3))));
+        }
+
+        return maxOps;
+    }
+}
+```
+
+#### Python
+```python
+class Solution:
+    def maxOperations(self, nums: List[int]) -> int:
+        from collections import defaultdict
+
+        n = len(nums)
+        score_count = defaultdict(int)
+
+        max_ops = 0
+        for i in range(n // 2):
+            score1 = nums[i] + nums[i + 1]
+            score2 = nums[n - 1 - i] + nums[n - 2 - i]
+            score3 = nums[i] + nums[n - 1 - i]
+
+            score_count[score1] += 1
+            score_count[score2] += 1
+            score_count[score3] += 1
+
+            max_ops = max(max_ops, score_count[score1], score_count[score2], score_count[score3])
+
+        return max_ops
+```
+