<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:
        score_1 = nums[0] + nums[1]
        score_2 = nums[-1] + nums[-2]
        score_3 = nums[0] + nums[-1]

        # print(score_1, score_2, score_3)
        # print(nums)
        # print(nums[1:-1])

        # use @cache to turn your recursive into memoization
        @cache
        def recursive(arr, t, curr_score):
            # print(arr, t)
            arr = tuple(arr)
            if len(arr) < 2:
                return 0

            score_1 = arr[0] + arr[1]
            score_2 = arr[-1] + arr[-2]
            score_3 = arr[0] + arr[-1]

            return_score = curr_score

            if score_1 == t:
                # print("score 1")
                return_score =  max(return_score, 1 + recursive(arr[2:], t, curr_score))

            if score_2 == t:
                # print("score 2")
                return_score =  max(return_score,  1 + recursive(arr[:-2], t, curr_score))

            if score_3 == t:
                # print("score 3")
                return_score =  max(return_score,  1 + recursive(arr[1:-1], t, curr_score))
            
            return return_score

        nums = tuple(nums)
        return max(recursive(nums, score_1, 0), recursive(nums, score_2, 0), recursive(nums, score_3, 0))

39821