<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:
        MAX_SUM = max(nums) * 2 + 1
        DIRS = [(2,0), (1,-1), (0,-2)]
        n = len(nums)

        if n < 2:
            return 0
        elif n == 2:
            return 1

        def search(s, e, target):
            if e - s <= 1:
                if s != e and nums[s] + nums[e] == target:
                    dp[s][e] = 1
                
                return dp[s][e]
            
            if dp[s][e] != 0:
                return dp[s][e]

            for ds, de in DIRS:
                ns, ne = s+ds, e+de

                if ns >= 0 and ns < n and ne >= 0 and ne < n and sum(nums[s:ns]) + sum(nums[ne+1:e+1]) == target:
                    dp[s][e] = max(dp[s][e], search(ns, ne, target) + 1)
            
            return dp[s][e]
        
        answer = 0
        s = 0
        e = len(nums) - 1
        answer = 0
        for ds, de in DIRS:
            dp = [[0] * n for _ in range(n)]
            ns, ne = s+ds, e+de
            if ns >= 0 and ns < n and ne >= 0 and ne < n:
                m = search(ns, ne, sum(nums[s:ns]) + sum(nums[ne+1:e+1])) + 1
                answer = max(answer, m)
        
        return answer

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