0377MediumArrayDynamic ProgrammingSliding Window

Longest Green Run After One Removal

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Trains the technique from

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

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A turnstile writes one entry per minute into log: 1 when it answered the gate controller that minute and 0 when it did not.

Before the log is filed, exactly one entry must be struck out -- the operator is required to remove one minute, not allowed to skip the removal. The auditor then reads the resulting array and reports the length of its longest run of consecutive 1s.

Return the largest length the auditor can be made to report. If no 1 can survive the removal, return 0.

Examples

Example 1

Input: log = [1, 1, 0, 1, 1, 1]
Output: 5

Striking out the entry at index 2 leaves five 1s in a row.

Example 2

Input: log = [1, 1, 1, 1]
Output: 3

Every minute answered, and one entry still has to go, so three remain in a row.

Example 3

Input: log = [1, 1, 0, 0, 1, 1, 1]
Output: 3

Striking out index 2 leaves [1, 1, 0, 1, 1, 1], whose longest run of 1s is three.

Example 4

Input: log = [0, 0, 0]
Output: 0

The log holds no 1 at all, so no run of 1s can survive.

Constraints

1 <= log.length <= 10^5
log[i] is either 0 or 1.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def longest_subarray(log: list[int]) -> int:

Java

public int longestSubarray(int[] log)

The intended approach

Time O(n)Space O(1)

Slide a window that is allowed to contain at most one `0`. Extend the right edge one minute at a time, and whenever a second `0` has entered, advance the left edge until only one is left inside. Every such window can be turned into a run of `1`s by striking out one of its entries, so the candidate answer is the window length minus one; the maximum of those candidates is the answer, and it is 0 for a log with no `1`.

Pythonentry point: longest_subarray

def longest_subarray(log):
    # Time O(n) -- each index enters and leaves the window once.
    # Space O(1) -- only counters.
    left = 0
    zeros = 0
    best = 0
    for right, entry in enumerate(log):
        if entry == 0:
            zeros += 1
        while zeros > 1:
            if log[left] == 0:
                zeros -= 1
            left += 1
        span = right - left + 1
        if span - 1 > best:
            best = span - 1
    return best

The obvious version

Try every index as the struck-out one, rebuild the shortened array, and scan it for its longest run of 1s. Quadratic, with no window reasoning at all.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_longest_subarray(log):
    best = 0
    for drop in range(len(log)):
        kept = log[:drop] + log[drop + 1:]
        run = 0
        for entry in kept:
            if entry == 1:
                run += 1
                if run > best:
                    best = run
            else:
                run = 0
    return best

How this problem was checked

Reference solution executed against 9 test cases — 9 passed
4 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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