<p>Given an array of integers, return the maximum sum for a <strong>non-empty</strong>&nbsp;subarray (contiguous elements) with at most one element deletion.&nbsp;In other words, you want to choose a subarray and optionally delete one element from it so that there is still at least one element left and the&nbsp;sum of the remaining elements is maximum possible.</p>

<p>Note that the subarray needs to be <strong>non-empty</strong> after deleting one element.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,-2,0,3]
<strong>Output:</strong> 4
<strong>Explanation: </strong>Because we can choose [1, -2, 0, 3] and drop -2, thus the subarray [1, 0, 3] becomes the maximum value.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,-2,-2,3]
<strong>Output:</strong> 3
<strong>Explanation: </strong>We just choose [3] and it&#39;s the maximum sum.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [-1,-1,-1,-1]
<strong>Output:</strong> -1
<strong>Explanation:</strong>&nbsp;The final subarray needs to be non-empty. You can&#39;t choose [-1] and delete -1 from it, then get an empty subarray to make the sum equals to 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= arr[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def maximumSum(self, arr: List[int]) -> int:
        check = max(arr)
        if check < 0:
            return check
        memo = {}
        
        def dp(idx, deleted):
            if idx == -1:
                return 0
            if (idx, deleted) not in memo:
                memo[(idx, deleted)] = arr[idx]
                if arr[idx] >= 0:
                    memo[(idx, deleted)] = max(memo[(idx, deleted)], arr[idx] + dp(idx-1, deleted))
                else:
                    if deleted:
                        memo[(idx, deleted)] = max(memo[(idx, deleted)], arr[idx] + dp(idx-1, deleted))
                    else:
                        memo[(idx, deleted)] = max(memo[(idx, deleted)], arr[idx] + dp(idx-1, deleted), dp(idx-1, True))
            return memo[(idx, deleted)]
        
        final = -inf
        for i in range(len(arr)):
            final = max(final, dp(i, False))

        return final



        
83000