<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 recursion(vector<int> &arr,int i,int t,int started,vector<vector<vector<int>>> &dp){
        if(i>=arr.size()) return 0;

        if(dp[t][started][i]!=-1) return dp[t][started][i];
        int ans =0;
        if(started){
            ans= max(ans,arr[i]+recursion(arr,i+1,t,started,dp));
            if(arr[i]<0 && t){
                ans= max(ans,recursion(arr,i+1,0,started,dp));
            }
        }
        else{
            ans= max(ans,recursion(arr,i+1,t,started,dp));
            ans= max(ans,arr[i]+recursion(arr,i+1,t,1,dp));
        }
        return dp[t][started][i]=ans;
    }
    int maximumSum(vector<int>& arr) {
        vector<vector<vector<int>>> dp(2,vector<vector<int>>(2,vector<int>(arr.size(),-1)));
        int ans=recursion(arr,0,1,0,dp);
        int maxi=INT_MIN;
        for(int i=0;i<arr.size();++i) maxi = max(maxi,arr[i]);
        if(ans==0) return maxi;
        return ans;
    }
};
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