<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:
vector<vector<long long>> memo;
long long dfs(vector<long long>& arr, int i, bool deleted) {
    if (i == arr.size()) return  INT_MIN;  // Safely terminate if deleted

    if (memo[i][deleted] != INT64_MIN) return memo[i][deleted];  // Check memoization

    // Option 1: Continue the subarray without deletion
    long long no_delete = max(arr[i], arr[i] + dfs(arr, i + 1, deleted));

    // Option 2: Delete the current element (only allowed if not already deleted)
    long long delete_current = deleted ? INT_MIN : dfs(arr, i + 1, true);

    // Return the best option between continuing without delete or with delete
    return memo[i][deleted] = max(no_delete, delete_current);
}

int maximumSum(vector<int>& arr) {
    int n = arr.size();
    memo.assign(n, vector<long long>(2, INT64_MIN));  // Initialize memoization table
    vector<long long> nn(n);
    for(int i = 0 ;i<n;i++){
        nn[i] = arr[i];
    }
    long long result = INT_MIN;
    for (int i = 0; i < n; ++i) {
        result = max(result, dfs(nn, i, false));  // Start from each index
    }
    return result;
}
};
49000