LeetCode 1493: Longest Subarray of 1's After Deleting One Element ·

Table of Contents

Meta Data
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Difficulty: medium First Attempt: 2025-08-24

Submissions: Accepted ×2

Approach
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A subarray that is all 1s after deleting one element was a subarray with at most one 0, and the deleted element is that 0 (or any 1 if there is no 0).

So find the longest window with at most one 0; the answer is window length − 1:

Add elements on the right; shrink the left while there is more than one 0.
Update with right - left (the length minus 1).

All 1s still require one deletion, giving n - 1, which this handles naturally.

My Solution
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class Solution:
    def longestSubarray(self, nums: List[int]) -> int:
        left = zeros = best = 0
        for right, x in enumerate(nums):
            zeros += x == 0
            while zeros > 1:
                zeros -= nums[left] == 0
                left += 1
            best = max(best, right - left)         # window length - 1
        return best

Both accepted versions were similar; the second had a cleaner while condition (the first also checked nums[right] == 0).

Complexity
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Time O(n), space O(1).

Mistakes to Avoid
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Returning the window length instead of length minus 1: one deletion is mandatory.
With all 1s the answer is n - 1, not n.

Interview Explanation
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Being all 1s after one deletion means the original subarray had at most one 0. I slide a window that allows at most one 0, shrinking the left when there are two. One element must be deleted, so the answer is the window length minus 1, which also covers the all-1s case. Time O(n).