<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:
        seq = []
        temp = 0
        cur_num = nums[0]
        for pos, i in enumerate(nums):
            if i == cur_num:
                temp += 1
            if i != cur_num:
                seq.append((cur_num, temp))
                temp = 1
                cur_num = i
        seq.append((cur_num, temp))
        seq_with_one = 0
        zero_total = 0
        deletable = 0
        for pos, s in enumerate(seq):
            if s[0] == 0:
                zero_total += s[1]
                if s[1] == 1 and pos != 0 and pos != len(seq) - 1:
                    deletable += 1
            if s[0] == 1:
                seq_with_one += 1
        print(seq_with_one, zero_total, deletable)
        if seq_with_one == 1 and zero_total != 0:
            return list(filter(lambda x: x[0] == 1, seq))[0][1]
        if seq_with_one == 1 and zero_total == 0:
            return list(filter(lambda x: x[0] == 1, seq))[0][1] - 1
        max_seq = 0
        for pos, s in enumerate(seq):
            if s == (0, 1) and pos != 0 and pos != len(seq) - 1:
                max_seq = max(max_seq, seq[pos - 1][1] + seq[pos + 1][1])
        for s in seq:
            max_seq = max(max_seq, s[1]) if s[0] == 1 else max_seq
        return max_seq
        
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