<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 reduce(nums, summ, i, j, dp):
            if dp.get(str(i)+","+str(j)) != None:
                return dp[str(i)+","+str(j)]
            if i+1>j:
                # return j-i
                # return len(nums) - (j-i)
                dp[str(i)+","+str(j)] = j-i+1
                return j-i+1

            temp1 = temp2 = temp3 = 100000000
            if nums[i]+nums[i+1] == summ[0]:
                temp1 = reduce(nums, summ, i+2, j, dp)
            if nums[i]+nums[j] == summ[0]:
                temp2 = reduce(nums, summ, i+1, j-1, dp)
            if nums[j]+nums[j-1] == summ[0]:
                temp3 = reduce(nums, summ, i, j-2, dp)
            
            if temp1 == temp2 and temp2 == temp3 and temp3 == 100000000:
                dp[str(i)+","+str(j)] = j-i+1
                return j-i+1
            dp[str(i)+","+str(j)] = min(temp1,temp2,temp3)
            return min(temp1,temp2,temp3)

        
        i = 0
        j = len(nums)-1
        dp = {}
        temp1 = reduce(nums, [nums[i]+nums[i+1]], i+2, j, dp)
        temp2 = reduce(nums, [nums[i]+nums[j]], i+1, j-1, dp)
        temp3 = reduce(nums, [nums[j]+nums[j-1]], i, j-2, dp)
        return (len(nums) - min(temp1,temp2,temp3))//2
        # ans = -1
        # if temp1 == None and temp2 == None and temp3 == None:
        #     return ans
        # elif temp1 != None:
        #     ans = temp1+1
        # elif temp2 != None:
        #     ans = temp2+1
        # else:
        #     ans = temp3+1
        # return len(nums)//2

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