<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) {
        vector <int> v;
        int temp=0,ans=0;
        for(int i=0;i<nums.size();i++){
            if(nums[i]==0){
                if(temp){
                    v.push_back(temp);
                }
                temp=0;
                v.push_back(0);
            }
            else{
                temp++;
            }
        }
        if(temp){
            v.push_back(temp);
        }
        if(v.size()==1){
            if(v[0]==0){
                return 0;
            }
            else{
                return v[0]-1;
            }
        }
        if(v.size()==2){
            if(v[0]==0){
                return v[1];
            }
            return v[0];
        }
        for(int i=0;i<v.size();i++){
            if(i-1<0 or i+1==v.size()){
                continue;
            }
            else{
                ans=max(ans,v[i-1]+v[i+1]);
            }
            ans=max(ans,v[i]);
        }
        return ans;
    }
};
61000