human
Solve in Python:
Given a binary array nums, you should delete one element from it.
Return the size of the longest non-empty subarray containing only 1's in the resulting array.
Return 0 if there is no such subarray.
 
Example 1:
Input: nums = [1,1,0,1]
Output: 3
Explanation: After deleting the number in position 2, [1,1,1] contains 3 numbers with value of 1's.
Example 2:
Input: nums = [0,1,1,1,0,1,1,0,1]
Output: 5
Explanation: After deleting the number in position 4, [0,1,1,1,1,1,0,1] longest subarray with value of 1's is [1,1,1,1,1].
Example 3:
Input: nums = [1,1,1]
Output: 2
Explanation: You must delete one element.
Example 4:
Input: nums = [1,1,0,0,1,1,1,0,1]
Output: 4

Example 5:
Input: nums = [0,0,0]
Output: 0

 
Constraints:

1 <= nums.length <= 10^5
nums[i] is either 0 or 1.
 Give me a the Python code you requested?
assistant
#Here is the Python code you requested.

class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        se=set(nums)
        if(0 not in se):
            return len(nums)-1
        elif(1 not in se):
            return 0
        
        count=0
        flag=0
        ls=[]
        for i in nums:
            if(i==0):
                if(flag==0):
                    count+=1
                else:
                    flag=0
                    ls.append(count)
                    count=1
            else:
                if(flag==1):
                    count+=1
                else:
                    flag=1
                    ls.append(count)
                    count=1
        ls.append(count)
        # print(ls)
        ind=0
        ans=0
        while(ind<len(ls)):
            if(ls[ind]!=0):
                if(ls[ind]==1):
                    if(ind==0 and ind<len(ls)-1):
                        ans=max(ans,ls[ind+1])
                    elif(ind==len(ls)-1 and ind>0):
                        ans=max(ans,ls[ind-1])
                    else:
                        if(ind<len(ls)-1 and ind>0):
                            ans=max(ans,ls[ind-1]+ls[ind+1])
                else:
                    if(ind==0 and ind<len(ls)-1):
                        ans=max(ans,ls[ind+1])
                    elif(ind==len(ls)-1 and ind>0):
                        ans=max(ans,ls[ind-1])
                    else:
                        if(ind<len(ls)-1 and ind>0):
                            ans=max(ans,max(ls[ind-1],ls[ind+1]))
            # print(ans,ind)
            ind+=2
            
        return ans