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.
class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        k = 1
        i = 0
        for j in range(len(nums)):
            if nums[j] == 0: k -= 1
            if k < 0:
                if nums[i] == 0: k += 1
                i += 1
        return j-i
class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        k = 1
        i = 0
        for j in range(len(nums)):
            if nums[j] == 0: k -= 1
            if k < 0:
                if nums[i] == 0: k += 1
                i += 1
        return j-i