<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 backtrack(left, right, target, memo):
            if (left, right) in memo:
                return memo[(left, right)]
            if left >= right:
                return 0
            answer = 0
            if (nums[left]+nums[right]) == target:
                answer = max(answer, 1+backtrack(left+1,right-1, target, memo))
            if (nums[left]+nums[left+1]) == target:
                answer = max(answer, 1+backtrack(left+2,right,target, memo))
            if (nums[right]+nums[right-1])==target:
                answer = max(answer, 1+backtrack(left,right-2, target, memo))
            
            memo[(left, right)] = answer
            return answer
        
        n = len(nums)-1
        one = backtrack(0, n, nums[0]+nums[n], {})
        two = backtrack(0, n, nums[0]+nums[1], {})
        three = backtrack(0, n, nums[-2]+nums[-1], {})
        return max(one, max(two, three))



189296