Editorial

Solution walkthrough

Review the explanation, then compare the reference implementation.

Video explanation

Solution

Intuition

Before deleting one element, a usable window contains at most one zero. Its length minus one is the number of ones that can remain: remove its zero if present, or remove a one when the window is already all ones. The deletion is mandatory.

Approach

Maintain a left boundary and a flag indicating whether the window contains a zero. When a second zero arrives, advance the left boundary past the previous zero, then retain the new zero in the window. After each right-end update, maximize right - left, which is the window length minus one. Return the best value; an all-ones array yields n - 1, not n.

class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        n = len(nums)
        z = False
        l = res = 0

        for r in range(n):
            if nums[r] == 0:
                while z:
                    if nums[l] == 0:
                        z = False
                    l += 1
                z = True
            res = max(res, r - l) # take maximum window - 1 | Always remove exactly 1 element

        return res

Complexity analysis

Each element enters and leaves the at-most-one-zero window once. Time is O(n) and auxiliary space is O(1).