<p>Given a binary array <code>nums</code>, you should delete one element from it.</p>

<p>Return <em>the size of the longest non-empty subarray containing only </em><code>1</code><em>&#39;s in the resulting array</em>. Return <code>0</code> if there is no such subarray.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,0,1]
<strong>Output:</strong> 3
<strong>Explanation:</strong> After deleting the number in position 2, [1,1,1] contains 3 numbers with value of 1&#39;s.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,1,1,1,0,1,1,0,1]
<strong>Output:</strong> 5
<strong>Explanation:</strong> After deleting the number in position 4, [0,1,1,1,1,1,0,1] longest subarray with value of 1&#39;s is [1,1,1,1,1].
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> You must delete one element.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>nums[i]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution {
public:
    int longestSubarray(vector<int>& nums) {
        long long int leftmax[nums.size()], rightmax[nums.size()];
       // memset(leftmax,0,sizeof(leftmax));
       // memset(rightmax,0,sizeof(rightmax));
        int n= nums.size();
        leftmax[0]= nums[0];
        rightmax[n-1] = nums[n-1];
        long long int maxans=0;
        if(n==1) return 0;
        for(int i=1;i<n;i++)
        {
            if(nums[i]!=0)
                leftmax[i] = nums[i] + leftmax[i-1];
            else
            leftmax[i]=0;    
        }
        for(int i=n-2;i>=0;i--)
        {
            if(nums[i]!=0)
                rightmax[i] = nums[i] + rightmax[i+1];
            else
            rightmax[i]=0;    
        }

        for(int i=0;i<n;i++)
        {
            if(i==0)
                maxans = max(maxans,rightmax[i+1]);
            else if(i==n-1)    
                maxans = max(maxans,leftmax[i-1]);
            else
            {
                maxans = max(maxans, leftmax[i-1]+rightmax[i+1]);
            }    

        }
        /*if(maxans == n-1)
        {
            maxans = maxans-1;
        }*/

        return maxans;
    }
};
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