<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 {
    const int fastio = [](){ ios::sync_with_stdio(false); cin.tie(nullptr); return 0; }();
public:
    int longestSubarray(vector<int>& nums) {
        const int nums_len = nums.size();

        deque<int> deq;
        int slow = 0;
        int maxl = 0;

        for (int fast = 0; fast < nums_len; ++fast) {
            if(nums[fast] == 0)
                deq.push_back(fast);
            
            if(deq.size() > 1) {
                maxl = max(maxl, fast - slow - 1);
                slow = 1 + deq.front();
                deq.pop_front();
            }
        }

        return max(maxl, nums_len - slow - 1);
    }
};
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