<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();
       vector<int> pre(n),suf(n),hash(n+1,0);
       for(int i=0;i<n;i++) hash[i+1]=hash[i]+arr[i];
       int sum=0,p=0,s=n;
       for(int i=1;i<=n;i++){   
            pre[i-1]=p;
            cout<<p<<' ';
            if(sum>hash[i]){
                sum=hash[i];
                p=i;
            }        
       } 
       sum=INT_MIN;
       for(int i=n;i>=1;i--){
            suf[i-1]=s;
            if(sum<hash[i]){
                sum=hash[i];
                s=i;
            }
       }
       cout<<endl;
       for(auto i:suf) cout<<i<<' ';
       cout<<endl;
       int maxm=INT_MIN;
       for(int i=0;i<n;i++){
            int s1=hash[i]-hash[pre[i]];
            int s2=hash[suf[i]]-hash[i+1];
            if(s1==0 && s2==0){
                maxm=max(maxm,arr[i]);
                continue;
            }
            maxm=max({maxm,s1+s2,arr[i],arr[i]+s1,arr[i]+s2});
       }
       return maxm;
    }
};
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