<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>& a) {
        int n=a.size();
        int left[n];
        int right[n];
        
        int sum=0 ,mx =INT_MIN;
        left[0]=a[0];sum=max(0,a[0]);
        int cm=a[0];
        for(int i=1;i<n;i++)
        {   
                cm =max(cm+a[i],a[i]);
                mx=max(mx,cm);
                sum+=a[i];
                mx= max(mx,sum);
                left[i]=sum;
                 if(sum<0)
                  sum=0;
               
             
        }
        sum=0;
        for(int i=n-1;i>=0;i--)
        {
                sum+=a[i];
                right[i]=sum;
                if(sum<0)
                sum=0;
               
               // mx= max(mx,right[i]);
        }

        mx= max(mx,left[n-1]);
        mx= max(mx,right[0]);

        for(int i=0;i<n;i++)
        {
            cout<<left[i]<<" ";
        }
        cout<<endl;
        for(int i=0;i<n;i++)
        {
            cout<<right[i]<<" ";
        }
        cout<<endl;
        
        for(int i=1;i<n-1;i++)
        {     
              mx=max(mx,left[i]);
              mx= max(mx,right[i]);
              if(a[i]<0)
              {
                mx=max(mx,left[i-1]);
                mx = max(mx,left[i-1]+right[i+1]);
              }
        }
        return mx;
    }
};
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