0778MediumArrayHash TableTwo PointersFloyd's Cycle Finding Algorithm

Same Way Round The Docking Ring

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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 cargo terminal has n docking bays laid out in a ring and numbered 0 to n - 1, with bay n - 1 sitting next to bay 0. Bay i carries a shunt value offsets[i], and a trolley standing at bay i is always shunted to bay (i + offsets[i]) mod n: a positive value pushes it that many bays forward round the ring, a negative value pulls it that many bays backward, and the ring wraps in both directions. A shunt value is never 0, and its size may be larger than n, in which case the trolley goes all the way round one or more times.

A round trip is a starting bay such that following the shunts from it eventually brings the trolley back to that same bay. A round trip counts only if both of these hold:

it touches two or more distinct bays, so a bay whose shunt lands on itself does not count; and
every shunt it uses points the same way, so either all of them are positive or all of them are negative.

Return true if at least one bay starts a round trip that counts, and false otherwise.

Examples

Example 1

Input: offsets = [1,6,2,1,4,1]
Output: true

Starting at bay 2 the trolley is shunted to bay 4, and bay 4 shunts it 4 forward which wraps back to bay 2. That round trip touches two distinct bays and both shunts point forward, so it counts.

Example 2

Input: offsets = [5,-5,5,-5]
Output: false

Bays 0 and 1 shunt to each other, and so do bays 2 and 3, but each of those round trips uses one forward shunt and one backward shunt, so neither counts.

Example 3

Input: offsets = [4,4,4,4]
Output: false

Every bay shunts the trolley a full lap round the ring and back onto itself, so each round trip here touches only one bay.

Constraints

1 <= offsets.length <= 5000
-1000 <= offsets[i] <= 1000
offsets[i] != 0

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 circular_array_loop(offsets: list[int]) -> bool:

Java

public boolean circularArrayLoop(int[] offsets)

The intended approach

Time O(n): every bay is retired at most once, and both pointers only ever move over bays that are still live.Space O(n) for the working copy of the shunt values.

Try every bay as a start, but walk it with two pointers that advance one and two shunts at a time and abandon the walk as soon as a shunt of the opposite sign appears. If the two pointers meet at a bay whose own shunt does not land on itself, that meeting point sits on a same-direction round trip of two or more bays. Either way the bays just walked cannot begin a qualifying round trip, so their shunts are zeroed out to keep later starts from re-walking them, which bounds the whole scan linearly.

Pythonentry point: circular_array_loop

def circular_array_loop(offsets):
    # Walk each start with a slow and a fast pointer, both stopping the moment a
    # move of the other sign shows up. Whatever the pair walked over can never
    # begin a qualifying circuit, so those bays are retired by zeroing them, and
    # every bay is therefore walked a bounded number of times.
    # Time O(n), space O(n) for the working copy of the offsets.
    step = list(offsets)
    n = len(step)

    def nxt(i):
        return (i + step[i]) % n

    for start in range(n):
        if step[start] == 0:
            continue
        sign = 1 if step[start] > 0 else -1
        slow = start
        fast = nxt(start)
        while step[fast] * sign > 0 and step[nxt(fast)] * sign > 0:
            if slow == fast:
                if slow != nxt(slow):
                    return True
                break
            slow = nxt(slow)
            fast = nxt(nxt(fast))
        i = start
        while step[i] * sign > 0:
            j = nxt(i)
            step[i] = 0
            i = j
    return False

The obvious version

From every bay, record the visit order of each bay reached until one repeats. The bays recorded at or after the repeat are exactly the round trip, so count how many of their shunts point forward and accept when the count is all or none, provided the round trip holds more than one bay. Quadratic in the number of bays.

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_circular_array_loop(offsets):
    n = len(offsets)
    for start in range(n):
        order = {}
        i = start
        t = 0
        while i not in order:
            order[i] = t
            t += 1
            i = (i + offsets[i]) % n
        ring = [bay for bay in order if order[bay] >= order[i]]
        if len(ring) > 1:
            forward = sum(1 for bay in ring if offsets[bay] > 0)
            if forward == 0 or forward == len(ring):
                return True
    return False

How this problem was checked

Reference solution executed against 17 test cases — 17 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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