<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:
    def longestSubarray(self, nums: List[int]) -> int:
        N = i = len(nums)
        prevK,k,maxK = 0, 0, 0
        try:
            i = nums.index(1)
            nums.append(0)
            nums.append(1)
        except:
            pass 
        
        remCnt = 0
        while i < N:
            remCnt += 1
            new_i = nums.index(0, i)
            k = new_i - i
            maxK = max(maxK, prevK + k)
            i = nums.index(1, new_i)
            prevK = k if i == (new_i + 1) else 0
        
        if remCnt == 1 and new_i == N and nums[0] != 0:
            maxK -= 1
        return maxK
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