<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:
        if len(set(nums)) == 1:
            return len(nums) // 2 
 
        n = len(nums)
        ans = 1
        
        @cache
        def helper(val, start, end, nops):
            if end - start < 1:
                return nops

            nops_max = nops
            firsttwo = nums[start] + nums[start + 1]
            lasttwo = nums[end - 1] + nums[end]
            firstlast = nums[start] + nums[end]
            if firsttwo == val:
                nops_max = max(nops_max, helper(val, start + 2, end, nops + 1))
            if lasttwo == val:
                nops_max = max(nops_max, helper(val, start, end - 2, nops + 1))
            if firstlast == val:
                nops_max = max(nops_max, helper(val, start + 1, end - 1, nops + 1))
            return nops_max

        ans = max(ans, helper(nums[0] + nums[1], 2, n - 1, 1))
        ans = max(ans, helper(nums[0] + nums[n - 1], 1, n - 2, 1))
        ans = max(ans, helper(nums[n - 2] + nums[n - 1], 0, n - 3, 1))
        return ans
19891