<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 dp[100001][2][2];
    
    int helper(vector<int>& arr, int idx, int start_exploring, int omitted_prev) {
        if(idx >= arr.size()) {
            return 0;
        }
        
        if(dp[idx][start_exploring][omitted_prev] != INT_MIN) {
            return dp[idx][start_exploring][omitted_prev];
        }
        
        int ans = INT_MIN+1;
        
        if(start_exploring) {
            ans = max(ans, arr[idx] + helper(arr, idx+1, 1, omitted_prev));
                
            if(omitted_prev) {
                ans = max(ans, 0);
            }
            else {
                ans = max(ans, helper(arr, idx+1, 1, 1));
            }
        }
        else {
            ans = max(ans, arr[idx] + helper(arr, idx+1, 1, omitted_prev));
            ans = max(ans, helper(arr, idx+1, 0, omitted_prev));
        }
        
        return dp[idx][start_exploring][omitted_prev] = ans;
    }
    
    int maximumSum(vector<int>& arr) {
        
        int max_ele = *max_element(arr.begin(), arr.end());
        if(max_ele <= 0) {
            return max_ele;
        }
        
        int n = arr.size();
        for(int i=0; i<=n; i++) {
            for(int j=0; j<2; j++) {
                for(int k=0; k<2; k++) {
                    dp[i][j][k] = INT_MIN;
                }
            }
        }
        return helper(arr, 0, 0, 0);
    }
};
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