<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 solve(vector<vector<int>>& dp,vector<int>& arr,int index,int n,int option)
    {
        if(index==n)
        return 0;
        if(dp[index][option]!=-1)
        return dp[index][option];
        int ans=0;
        if(option==0)
        {
            if(index!=n-1)
            ans=max(solve(dp,arr,index+1,n,option),arr[index]+solve(dp,arr,index+1,n,1));
            else
            ans=arr[index]+solve(dp,arr,index+1,n,1);
        }
        else if(option==1)
        {
            ans=max(arr[index]+solve(dp,arr,index+1,n,1),solve(dp,arr,index+1,n,2));
        }
        else if(option==2)
        {
            ans=max(arr[index]+solve(dp,arr,index+1,n,2),solve(dp,arr,index+1,n,3));
        }
        else
        {
            ans=solve(dp,arr,index+1,n,3);
        }
        return dp[index][option]=ans;
    }
    int maximumSum(vector<int>& arr) {
       int n=arr.size();
       vector<vector<int>> dp(n,vector<int>(4,-1));
       return solve(dp,arr,0,n,0); 
    }
};
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