<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 {
public:
    int maximumSum(vector<int>& arr) {

        int n = arr.size(); 
        int _max = arr[0]; 
        bool check = true; 
        for( int i = 0; i < n; i++ )
        {
            if( arr[i] >= 0 )
            {
                check = false; 
                break;
            }
            _max = max(_max, arr[i]); 
        }

        if( check )
            return _max; 

        vector<int> pref(n , 0); 
        vector<int> min_pref(n , 0); 
        vector<int> suff(n , 0); 
        vector<int> min_suff(n , 0); 
        
        int minp = 0; 
        pref[0] = arr[0];
        minp = min( minp, pref[0] );

        for( int i = 1; i < n; i++ )
        {
            min_pref[i] = minp; 
            pref[i] = pref[i-1] + arr[i]; 
            minp = min( minp, pref[i] ); 
        }

        int mins = 0; 
        suff[n-1] = arr[n-1]; 
        mins = min( mins, suff[n-1]); 

        for( int i = n-2; i >= 0; i-- )
        {
            min_suff[i] = mins; 
            suff[i] = suff[i+1] + arr[i]; 
            mins = min( mins, suff[i]); 
        }

        int ans = 0; 
        for( int i = 0; i < n; i++ )
        {
            if( arr[i] < 0 )
            ans = max( ans , pref[i] - min_pref[i] + suff[i] - min_suff[i] - 2*arr[i] ); 
            else 
            ans = max( ans , pref[i] - min_pref[i] + suff[i] - min_suff[i] - arr[i] ); 
        }

        return ans; 
    }
};
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