<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) {
        ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
        int n = nums.size();
        if(n == 1) return 0;

        int ans = INT_MIN;
        int l=0, r=0;
        int zero_cnt = 0; //try to maintain window which has at most one zero!
        
        while(r<n && l <= r){
            if(nums[r] == 0){
                zero_cnt++;
            }

            if(zero_cnt > 1){
                //shrink
                if(zero_cnt > 1){
                    if(nums[l] == 0){
                        zero_cnt--;
                    }
                    l++;
                }
            }

            ans = max(ans, r-l);
            r++;
        }

        return ans;
    }
};
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