0022MediumArrayHash TableUnion-Find

Widest Serviced Floor Band

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Trains the technique from

LeetCode 128Longest Consecutive Sequence

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 building automation system writes one entry every time a lift stops to be serviced. floors[i] is the floor number attached to entry i. Floors above ground carry positive numbers, ground is 0, and basement levels carry negative numbers. Entries land in whatever order the technicians filed them, and the same floor may show up several times.

Call a set of floor numbers a band when it has no holes: some floor f, then f + 1, then f + 2, and so on up to some final floor. A band counts as serviced when every floor it contains shows up at least once somewhere in floors. Where those entries sit in the log is irrelevant.

Return how many floors the widest serviced band contains. An empty log services no band at all, so the answer is 0 in that case.

Repeats never widen a band: three entries for floor 4 still only account for one floor.

The log may hold up to a hundred thousand entries, and your routine has to finish in expected linear time with respect to that count. Putting the log in order first is therefore ruled out.

Examples

Example 1

Input: floors = [3, 12, 4, 2, 11, 5]
Output: 4

Floors 2, 3, 4 and 5 are all present, which is a band of four floors. Adding 6 would break it because no entry mentions floor 6, and the pair 11 with 12 only reaches two floors.

Example 2

Input: floors = [6, 7, 7, 8, 8, 12]
Output: 3

The band 6 through 8 is serviced, so the answer is three. The duplicate entries for 7 and 8 add nothing, and floor 12 stands alone because floors 9 through 11 never appear.

Example 3

Input: floors = [-2, 9, -1, 0, -3]
Output: 4

Three basement levels join ground level to form the band -3 through 0, four floors wide. Floor 9 is isolated since neither 8 nor 10 was logged.

Constraints

0 <= floors.length <= 10^5
-10^9 <= floors[i] <= 10^9
Expected running time is linear in floors.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 widest_floor_band(floors: list[int]) -> int:

Java

public int widestFloorBand(int[] floors)

The intended approach

Time O(n) expectedSpace O(n)

Drop every entry into a hash set, which erases duplicates and answers membership questions in constant time. A floor starts a band only when the floor one below it is missing from the set, so walk upward from each such starting floor and count how far the chain continues. Every floor is visited at most twice overall, once as a candidate start and once while climbing a single chain, which keeps the whole pass linear in expectation.

Pythonentry point: widest_floor_band

def widest_floor_band(floors):
    present = set(floors)
    widest = 0
    for value in present:
        if value - 1 in present:
            continue
        span = 1
        nxt = value + 1
        while nxt in present:
            span += 1
            nxt += 1
        if span > widest:
            widest = span
    return widest

The obvious version

Deduplicate, sort ascending, then sweep once counting how long each unbroken step-of-one run lasts. Sorting is allowed here because only correctness matters for the reference check.

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_widest_band(floors):
    ordered = sorted(set(floors))
    if not ordered:
        return 0
    widest = 1
    run = 1
    for i in range(1, len(ordered)):
        if ordered[i] == ordered[i - 1] + 1:
            run += 1
        else:
            run = 1
        if run > widest:
            widest = run
    return widest

How this problem was checked

Reference solution executed against 12 test cases — 12 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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