0944MediumTwo PointersStringGreedy

Widest Step Between Assigned Benches

Tracked in this browser only

Trains the technique from

LeetCode 4026Maximum Gap Between Stations

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.

Workers are lined up in the order given by skill, one letter each, and benches are lined up in the order given by station, one letter each. There are at least as many benches as workers.

Each worker is put on a bench whose letter matches their own, one worker per bench, and the workers keep their order: an earlier worker must take an earlier bench. At least one such assignment is guaranteed to exist.

The step between two neighbouring workers is the difference between their bench positions. Return the widest step any assignment can have, or 0 when there is only one worker.

Examples

Example 1

Input: skill = "ab", station = "aabb"
Output: 3

The worker wanting an a can take bench 0 and the one wanting a b can take bench 3, which is as far apart as they go.

Example 2

Input: skill = "a", station = "a"
Output: 0

With a single worker there is no neighbouring pair to measure.

Example 3

Input: skill = "ab", station = "ab"
Output: 1

Each worker has exactly one bench that matches, so the step is fixed at one.

Constraints

1 <= skill.length <= station.length <= 10^5
skill and station consist of lowercase English letters only
At least one assignment exists

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 maximum_gap(skill: str, station: str) -> int:

Java

public int maximumGap(String skill, String station)

The intended approach

Time O(n + m) for n workers and m benches: two passes with a pointer walking the benches.Space O(n) for the two bench lists.

The widest step is between one particular neighbouring pair, so work out for each pair how far apart they can be pushed and take the largest. Pushing a pair apart means putting the earlier worker as far left as possible and the later one as far right as possible, and the two demands do not conflict, because everything before the earlier worker can be crammed to the left and everything after the later one crammed to the right. So make two passes. Going left to right, greedily give each worker the earliest bench that matches and comes after the previous worker's bench; that gives each worker their earliest possible bench. Going right to left the same way gives each worker their latest. The answer is the largest, over neighbouring pairs, of the later worker's latest bench less the earlier worker's earliest.

Pythonentry point: maximum_gap

def maximum_gap(skill, station):
    n = len(skill)
    m = len(station)
    earliest = [0] * n
    at = 0
    for i in range(n):
        while at < m and station[at] != skill[i]:
            at += 1
        earliest[i] = at
        at += 1
    latest = [0] * n
    at = m - 1
    for i in range(n - 1, -1, -1):
        while at >= 0 and station[at] != skill[i]:
            at -= 1
        latest[i] = at
        at -= 1
    best = 0
    for i in range(1, n):
        step = latest[i] - earliest[i - 1]
        if step > best:
            best = step
    return best

The obvious version

Walk out every assignment there is, picking a matching bench for each worker in turn and always moving rightwards, and keep the widest neighbouring step over all of them. Exponential in the bench count and it never reasons about pushing a pair apart.

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_maximum_gap(skill, station):
    n = len(skill)
    m = len(station)
    best = [0]

    def walk(worker, from_bench, chosen):
        if worker == n:
            widest = 0
            for i in range(1, n):
                step = chosen[i] - chosen[i - 1]
                if step > widest:
                    widest = step
            if widest > best[0]:
                best[0] = widest
            return
        for bench in range(from_bench, m):
            if station[bench] != skill[worker]:
                continue
            chosen.append(bench)
            walk(worker + 1, bench + 1, chosen)
            chosen.pop()

    walk(0, 0, [])
    return best[0]

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

Create a free account to unlock the solution

See the step-by-step animation, the intuition, and clean code in every language — free, no credit card.

Create a free account Already have an account? Sign in