<p>Given an array of integers called <code>nums</code>, you can perform <strong>any</strong> of the following operation while <code>nums</code> contains <strong>at least</strong> <code>2</code> elements:</p>

<ul>
	<li>Choose the first two elements of <code>nums</code> and delete them.</li>
	<li>Choose the last two elements of <code>nums</code> and delete them.</li>
	<li>Choose the first and the last elements of <code>nums</code> and delete them.</li>
</ul>

<p>The<strong> score</strong> of the operation is the sum of the deleted elements.</p>

<p>Your task is to find the <strong>maximum</strong> number of operations that can be performed, such that <strong>all operations have the same score</strong>.</p>

<p>Return <em>the <strong>maximum</strong> number of operations possible that satisfy the condition mentioned above</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,2,1,2,3,4]
<strong>Output:</strong> 3
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,2,6,1,4]
<strong>Output:</strong> 2
<strong>Explanation:</strong> We perform the following operations:
- Delete the first two elements, with score 3 + 2 = 5, nums = [6,1,4].
- Delete the last two elements, with score 1 + 4 = 5, nums = [6].
It can be proven that we can perform at most 2 operations.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 2000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
</ul>

class Solution:
    def maxOperations(self, nums: List[int]) -> int:
        def helper(A, start, end, prev_sum, count, memo):
            if (start, end, prev_sum) in memo:
                return memo[(start, end, prev_sum)]

            if (end - start + 1) < 2:
                return count

            max_count = count

            # Option 1: Remove first two elements
            if start + 1 <= end:
                sum1 = A[start] + A[start + 1]
                if sum1 == prev_sum or prev_sum is None:
                    max_count = max(max_count, helper(A, start + 2, end, sum1, count + 1, memo))

            # Option 2: Remove last two elements
            if end - 1 >= start:
                sum2 = A[end] + A[end - 1]
                if sum2 == prev_sum or prev_sum is None:
                    max_count = max(max_count, helper(A, start, end - 2, sum2, count + 1, memo))

            # Option 3: Remove first and last elements
            if start < end:
                sum3 = A[start] + A[end]
                if sum3 == prev_sum or prev_sum is None:
                    max_count = max(max_count, helper(A, start + 1, end - 1, sum3, count + 1, memo))

            memo[(start, end, prev_sum)] = max_count
            return max_count
        
        memo = {}
        return helper(nums, 0, len(nums) - 1, None, 0, memo)
        
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