<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>


def recur_dp(nums, target, i, j, vis):
    if (i, j) in vis:
        return vis[(i, j)]

    if j - i < 1:
        return 0
    
    acc = []
    # left-most
    if nums[i] + nums[i+1] == target:
        acc.append(recur_dp(nums, target, i+2, j, vis) + 1)
    # right-most
    if nums[j-1] + nums[j] == target:
        acc.append(recur_dp(nums, target, i, j-2, vis) + 1)
    # first and last
    if nums[i] + nums[j] == target:
        acc.append(recur_dp(nums, target, i+1, j-1, vis) + 1)

    vis[(i, j)] = max(acc) if len(acc) else 0
    # print(i,j,vis[(i,j)])
    return vis[(i, j)]
    


def sub_dp(nums, target):
    # print(f"{target=}")
    vis = {}
    return recur_dp(nums, target, 0, len(nums) - 1, vis)


class Solution:
    def maxOperations(self, nums: List[int]) -> int:
        acc = [
            1 + sub_dp(nums[2:], nums[0] + nums[1]),
            1 + sub_dp(nums[:-2], nums[-2] + nums[-1]),
            1 + sub_dp(nums[1:-1], nums[0] + nums[-1]),
        ]
        return max(acc)
219191