<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 = {}
        
        def recurse(l, r, prev):

            if (l, r, prev) in memo:
                return memo[(l, r, prev)]
            
            if l >= r:
                return 0
            
            res1, res2, res3 = 0, 0, 0
            
            # Delete first two
            curr_score1 = nums[l] + nums[l + 1]
            if curr_score1 == prev or prev == None:
                res1 = 1 + recurse(l + 2, r, curr_score1)

            # Delete last two
            curr_score2 = nums[r] + nums[r - 1]
            if curr_score2 == prev or prev == None:
                res2 = 1 + recurse(l, r - 2, curr_score2)
            
            # Delete first and last
            curr_score3 = nums[l] + nums[r]
            if curr_score3 == prev or prev == None:
                res3 = 1 + recurse(l + 1, r - 1, curr_score3)

            res = max(res1, res2, res3)
                
            memo[(l, r, prev)] = res
                
            return res
        
        return recurse(0, len(nums) - 1, None)
398561