<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 f(int i, int j, vector<vector<int>>& dp, vector<int>& arr) {
        // Base case: if i exceeds array size
        if (i == arr.size()) 
            return 0; // A very negative number since we are summing

        // If the value is already computed, return it
        if (dp[i][j] != -1) 
            return dp[i][j];

        // Option 1: Take the current element
        int take = arr[i] + f(i + 1, j, dp, arr);

        // Option 2: Skip the current element if j == 1 (meaning we can still delete one element)
        int skip = 0; // Cannot skip anymore if j == 0
        if (j == 1)
            skip = f(i + 1, 0, dp, arr);

        // Store and return the maximum of taking or skipping the element
        return dp[i][j] = max(take, skip);
    }

    int maximumSum(vector<int>& arr) {
        int n = arr.size();

        // Initialize dp array to store results
        vector<vector<int>> dp(n, vector<int>(2, -1));
        
        int ans=-1e9;
        bool fl=false;
        for(int i=0;i<arr.size();i++)
        {    if(arr[i]>=0)fl=true;
        ans=max(ans,arr[i]);
        }
        if(!fl)
            return ans;
        // Compute the maximum sum where exactly one element is deleted (start from index 0)
        int res=-1;
        for(int i=0;i<arr.size();i++){
            res=max(res,f(i,1,dp,arr));
        }
return res;
        
    }
};

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