<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:
        n = len(arr)
        if n ==1:
            return arr[0]
        dpLeft = [-10**4-1 for _ in range(n)]
        dpLeft[0] = arr[0]
        i = 1 
        while i < n :
            dpLeft[i] = max(arr[i],dpLeft[i-1]+arr[i])
            i += 1
        dpRight = [-10**4-1 for _ in range(n)]
        dpRight[-1] = arr[-1]
        j = n-2 
        while j >= 0:
            dpRight[j] = max(arr[j],dpRight[j+1]+arr[j])
            j -= 1
        k = 0
        maxi = -10**4-1
        while k < n:
            # if we take it 
            curr_maxi_with = dpRight[k] + dpLeft[k] - arr[k]
            
            if k==0:
                curr_maxi_without = dpRight[k+1]
            elif k==n-1:
                curr_maxi_without = dpLeft[k-1]
            else:
                if dpLeft[k-1]>=0 and dpRight[k+1]>=0:
                    curr_maxi_without = dpRight[k+1] + dpLeft[k-1]
                else:
                    curr_maxi_without = max(dpLeft[k-1],dpRight[k+1])
            
            maxi= max(maxi,curr_maxi_without, curr_maxi_with)
            k += 1
        
        return maxi
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