0058MediumArrayBinary SearchSliding WindowPrefix Sum

Longest Verified Backup Streak

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Trains the technique from

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 storage team keeps a nightly integrity log. checks holds one entry per night in chronological order: 1 means that night's snapshot was checked, 0 means the check was skipped.

There is spare capacity to go back and check at most k of the skipped nights, and the team may choose whichever skipped nights it likes.

Report the greatest number of neighbouring nights that can all end up checked once at most k retro-checks have been spent. Spare capacity may go unused, and the nights picked for a retro-check need not sit next to each other.

Examples

Example 1

Input: checks = [1, 1, 0, 1, 1, 0, 1], k = 1
Output: 5

Spending the single retro-check on night 2 leaves nights 0 through 4 all checked, five nights wide. Any six neighbouring nights here hold two gaps, which is one retro-check too many.

Example 2

Input: checks = [0, 0, 1, 1, 0, 0, 1, 1, 1, 0], k = 2
Output: 7

Retro-checking nights 4 and 5 fills the only gaps inside nights 2 through 8, so seven neighbouring nights end up checked.

Example 3

Input: checks = [0, 0, 0], k = 0
Output: 0

Nothing was checked and there is no spare capacity, so no night can be part of a checked block.

Constraints

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

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_verified_streak(checks: list[int], k: int) -> int:

Java

public int longestVerifiedStreak(int[] checks, int k)

The intended approach

Time O(n)Space O(1)

A block of nights is reachable exactly when the number of zeros inside it is within `k`, so slide a window and keep a count of the zeros it contains. Push the right edge forward one night at a time and pull the left edge forward while the zero count exceeds `k`, releasing a zero from the count each time the left edge steps over one. The widest window ever held is the answer, and each edge only ever moves forward.

Pythonentry point: longest_verified_streak

def longest_verified_streak(checks, k):
    left = 0
    skipped = 0
    best = 0
    for right in range(len(checks)):
        if checks[right] == 0:
            skipped += 1
        while skipped > k:
            if checks[left] == 0:
                skipped -= 1
            left += 1
        width = right - left + 1
        if width > best:
            best = width
    return best

The obvious version

Anchor a start at every night, walk forward while tallying skipped nights, and stop that walk as soon as the tally passes k. O(n^2).

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_verified_streak(checks, k):
    n = len(checks)
    best = 0
    for start in range(n):
        skipped = 0
        for end in range(start, n):
            if checks[end] == 0:
                skipped += 1
            if skipped > k:
                break
            width = end - start + 1
            if width > best:
                best = width
    return best

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
3 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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