<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:
        memo = [[0] * len(nums) for _ in range(len(nums))]
        def dfs(l, r, total):
            if l >= r:
                return 0
            if memo[l][r] != 0:
                return memo[l][r]
            maxOps = 0
            if nums[l] + nums[l+1] == total:
                maxOps = max(maxOps, 1 + dfs(l+2, r, total))
            if nums[r] + nums[r-1] == total:
                maxOps = max(maxOps, 1 + dfs(l, r-2, total))
            if nums[l] + nums[r] == total:
                maxOps = max(maxOps, 1 + dfs(l+1, r-1, total))
            memo[l][r] = maxOps
            return maxOps

        l, r = 0, len(nums)-1
        res = 0
        res = 1 + max(res, dfs(l+2, r, nums[l] + nums[l+1]), dfs(l, r-2, nums[r]+nums[r-1]), dfs(l+1, r-1, nums[l]+nums[r]))
        return res

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