0183EasyArray

Longest Online Panel Run

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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 solar farm polls its panels along one long row and hands you the reply as status, listed in the order the panels sit on the row. status[i] is 1 when panel i answered and 0 when it did not.

A crew wants to know the widest stretch of neighbouring panels that all answered. Return the number of panels in the longest stretch that contains no silent panel, or 0 when no panel answered at all.

Examples

Example 1

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

The first three panels answer in a row. The pair at the end is only two wide.

Example 2

Input: status = [0, 1, 0, 0, 1, 1, 1, 1]
Output: 4

The stretch runs to the end of the row, so it still counts as four panels.

Example 3

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

Nothing answered, so there is no stretch to measure.

Constraints

1 <= status.length <= 10^5
status[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_online_run(status: list[int]) -> int:

Java

public int longestOnlineRun(int[] status)

The intended approach

Time O(n)Space O(1)

Sweep the row once. Grow a counter on every panel that answered and compare it against the best seen right away, so a stretch that runs off the end of the row is already recorded. A silent panel drops the counter back to zero.

Pythonentry point: longest_online_run

def longest_online_run(status):
    # Single sweep holding the run in progress and the best run seen.
    # Time O(n), space O(1).
    best = 0
    run = 0
    for flag in status:
        if flag == 1:
            run += 1
            if run > best:
                best = run
        else:
            run = 0
    return best

The obvious version

Start a stretch at every panel and extend it forward until a silent panel stops it, keeping the widest measurement. 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_online_run(status):
    best = 0
    for start in range(len(status)):
        length = 0
        for i in range(start, len(status)):
            if status[i] != 1:
                break
            length += 1
        if length > best:
            best = length
    return best

How this problem was checked

Reference solution executed against 9 test cases — 9 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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