<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:

        dp = {}
        dp[nums[0] + nums[1]] = defaultdict(dict)
        dp[nums[0] + nums[-1]] = defaultdict(dict)
        dp[nums[-2] + nums[-1]] = defaultdict(dict)

        def dfs(i, j, score):
            if len(nums[i:j]) < 2:
                return 0
            if i in dp[score] and j in dp[score][i]:
                return dp[score][i][j]
            
            cond1 = nums[i] + nums[i + 1] == score
            cond2 = nums[i] + nums[j - 1] == score
            cond3 = nums[j - 2] + nums[j - 1] == score

            ans1 = dfs(i + 2, j, score) if cond1 else 0
            ans2 = dfs(i + 1, j - 1, score) if cond2 else 0
            ans3 = dfs(i, j - 2, score) if cond3 else 0
            
            if not (cond1 or cond2 or cond3):
                dp[score][i][j] = 0
            else:
                dp[score][i][j] = max(ans1, ans2, ans3) + 1

            return dp[score][i][j]

        for score in dp.keys():
            dfs(0, len(nums), score)

        max_numbers = [item[0][len(nums)] for score, item in dp.items()]
        return max(max_numbers)
        
169366