<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 ans=INT_MIN;
    int dp[2][100005];
    int sum(int i,vector<int>&arr,int skip=0)
    {
        if(i>=arr.size()) return 0;
        if(dp[skip][i]!=-1) return dp[skip][i];
        int op1=0,op2=INT_MIN;
        op1=arr[i];
        int k=sum(i+1,arr,skip);
        if(k>0)
        {
            op1+=k;
        }

        if(arr[i]<0 && skip==0)
        {
            
            op2=sum(i+1,arr,1);
        }
        // cout<<i<<" "<<skip<<" "<<op1<<" "<<op2<<endl;
        ans=max(ans,op1);
        if(i!=arr.size()-1)
            ans=max(ans,op2);

        return  dp[skip][i]=max(op1,op2);
    }
    int maximumSum(vector<int>& arr) {
        ans=INT_MIN;
        memset(dp,-1,sizeof(dp));
        sum(0,arr,0);
        return ans;
    }
};
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