<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>

"""
Clarifications:
    binary number, delete one element from it
    only one's in array
    You can only delete one 1
    nums = [1,1,0,1] => [1, 1, 1]
    nums = [0,1,1,1,0,1,1,0,1] => 5

Approach 1: TC:O(N^2) | SC:O(1)
    start from each loop, and count every and put count 

Approach 2: TC:O(N) | SC:O(1)
    put a counter, whenever, count == 2, increase l

Edge case:
    can nums be empty? No
    only 0 and 1? Yes
"""
class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        res = 0
        l = 0
        count_zeros = 0
        flag_found_zero = False
        for r in range(len(nums)):
            if nums[r] == 0:
                count_zeros += 1
                flag_found_zero = True
            while count_zeros >= 2:
                if nums[l] == 0:
                    count_zeros -= 1
                l += 1
            res = max(res, r - l + 1 - count_zeros)
        return res if  flag_found_zero else res - 1
"""
Test cases:
            0   1   2   3   4   5   6   7   8
    nums = [0,  1,  1,  1,  0,  1,  1,  0,  1]
                                        i
                l
    res = 3.  #4 - 1 + 1 - 1
    l = 0
    count_zeros = 1
"""

        
492