<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>

const int INF = 1e9 + 7;
class Solution {
public:
    int mx = -INF;
    int maximumSum(vector<int>& arr) {
        int n = arr.size();
        int dp[n][2];
        memset(dp, -1, sizeof(dp));
        func(n - 1, 1, n, arr, dp);
        return mx;
    }

    int func(int idx, int k, int &n, vector<int>& arr, int dp[][2]) {
        if(idx == 0) {
            mx = max(mx, arr[0]);
            return arr[0];
        }

        if(dp[idx][k] != -1) return dp[idx][k];

        // start a new subarray
        int val = arr[idx];

        // do not delete element at idx
        val = max(val, func(idx - 1, k, n,arr, dp) + arr[idx]);

        if(k) {
            // deletions are left therefore delete element at k
            val = max(val, func(idx - 1, k - 1, n, arr, dp));
        }

        mx = max(mx, val);
        return dp[idx][k] = val;
    }
};

/*
    func(idx, k) -> The maximum subarray ending at idx with k deletions

    at idx:
        do not delete idx -> func(idx - 1, k)
        start a new subarray which ends at idx -> arr[idx]

        delete idx:
            func(idx - 1, k - 1)
    
*/
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