Longest subarray of 1s after deleting one element

class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        l = r = 0
        candel = 1
        res = 0
        while r < len(nums):
            if nums[r] == 0:
                candel -= 1
            if candel < 0:
                if nums[l] == 0:
                    candel += 1
                l += 1
            r += 1
            res = max(res, r - l - 1)
        return res

Longest Subarray of 1's After Deleting One Element

Difficulty: Medium

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.

 

Constraints:

  • 1 <= nums.length <= 105
  • nums[i] is either 0 or 1.