<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 {
    int solve(int i, int k, vector<int>& arr, vector<vector<int>>& dp, int& ans)
    {   //base case
        if(i == arr.size())
         return -1e9;

        //memoization
        if(dp[i][k] != -1)
         return dp[i][k];

        int take = INT_MIN, notTake = INT_MIN;
        //we can always take current element
         dp[i][k] = max(arr[i], arr[i] + solve(i+1,k,arr,dp,ans));
      if(k==1) //this means we can delete currunt element 
         dp[i][k] = max(dp[i][k], solve(i+1,0,arr,dp,ans)); //now k is set to 0, means deletion not allowed again

        ans = max(ans,dp[i][k]);
        return dp[i][k];
    }
public:
    int maximumSum(vector<int>& arr) {
        int maxi = 0, n = arr.size();;
        vector<vector<int>> dp(n,vector<int>(2,-1));
        int ans = arr[0];
        int x = solve(0,1,arr,dp,ans);
        return ans;
    }
};
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