<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 i, int n, vector<int>&v, int skip, vector<vector<int>>&dp)
    {
        if(skip == 3) return 0;
        if(i == n) 
        {
            if(skip == 0) return -1e8;
            return 0;
        }
        if(dp[i][skip] != -1) return dp[i][skip];

        int nt = -1e8, t = -1e8;
        if(skip == 0)
        {
            nt = ans(i + 1, n, v, skip, dp);
        }
        else
        {
            nt = ans(i + 1, n, v, skip + 1, dp);
        }
        t = v[i] + ans(i + 1, n, v, max(skip, 1), dp);

        return dp[i][skip] = max(t, nt);
    }

    int maximumSum(vector<int>& arr) {
        int n = arr.size();
        vector<vector<int>>dp(n, vector<int>(4, -1));
        return ans(0, n, arr, 0, dp);
    }
};
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