<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) {

       int ans=0;
        int i=0;
        int j=0;
        int zero=0;

        vector<int> vec;

        int temp=vec.size();
        int k=1;

        while(i<nums.size()){
            if(nums[i]==0){
                zero++;
            }

            if(zero<=k){
                vec.push_back(nums[i]);
                i++;
                ans = max(ans, static_cast<int>(vec.size()));

            }
            else if(zero>k){
                ans = max(ans, static_cast<int>(vec.size()));

                while( j<nums.size() && nums[j]!=0){
                    j++;
                }
                j++;
                vec.clear();

                zero=0;
                i=j;
            }

            
        }

       

        return ans-1;
    }
};
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