<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:
        prev = None
        k = 1
        count = 0
        res = []
        has_zeros = False
        for i in nums:
            if prev is None:
                prev = i
                count = 1
            elif prev == i:
                count += 1
            else:
                if prev == 1:
                    res.append(count)
                else:
                    has_zeros = True
                    res.append(-count)
                prev = i
                count = 1
        if prev == 1:
            res.append(count)
        else:
            res.append(-count)
            has_zeros = True
        mx = 0
        for i in range(len(res)):
            if i<0:
                continue
            else:
                m =res[i]
                j = i+1
                buffer = k
                while buffer>= 0 and j<len(res):
                    if res[j]>0:
                        m+=res[j]
                    else:
                        buffer+=res[j]
                    j+=1
                mx = max(mx,m)
        mx = mx if has_zeros else mx-1
        mx = mx if mx > 0 else 0
        return min(mx, len(nums))
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