<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:
        zero_indices = []
        if nums == []: 
            return 0
        if 1 not in nums: 
            return 0
        if 0 not in nums: 
            return len(nums)-1
        for i in range(len(nums)): 
            if i == 0 and nums[i] != 0: 
                zero_indices.append(-1)
            if i == len(nums)-1 and nums[-1] != 0: 
                zero_indices.append(len(nums))
            if nums[i] == 0: 
                zero_indices.append(i)
        
        diffs = []
        for i in range(len(zero_indices)): 
            if i == 0: 
                diffs.append(zero_indices[1] - zero_indices[0] - 1)
            elif i == len(zero_indices)-1: 
                diffs.append(zero_indices[-1] - zero_indices[-2] - 1)
            else: 
                diffs.append((zero_indices[i] - zero_indices[i-1] - 1) + (zero_indices[i+1]-zero_indices[i]-1))
        
        max_num = max(diffs)
        return max_num

            
                
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