0476MediumArrayHash TableBinary Search

Nearest Twin Bay on the Carousel

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Trains the technique from

LeetCode 3488Closest Equal Element Queries

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 picking carousel has n storage bays bolted round a closed ring, numbered 0 to n - 1 in the order they sit on the ring, with bay n - 1 sitting right next to bay 0. Bay i holds the product coded codes[i]; the same code may be stocked in several bays.

One step of the carousel moves you from a bay to either of the two bays beside it, so it can be turned in either direction.

For every entry in asks, treat asks[j] as a starting bay and report the fewest steps from it to a different bay holding the same code. Report -1 for that entry when no other bay holds that code. Because a step count is never negative, -1 cannot be confused with a real answer.

Return the answers as a list in the same order as asks.

Examples

Example 1

Input: codes = [7, 2, 9, 2, 7], asks = [0, 1, 2]
Output: [1, 2, -1]

Bays 0 and 4 both hold code 7 and sit next to each other across the seam of the ring. Bays 1 and 3 both hold code 2. Code 9 is stocked once only.

Example 2

Input: codes = [7, 3, 3, 3, 7, 3, 3, 7], asks = [0, 7, 4]
Output: [1, 1, 3]

Code 7 sits in bays 0, 4 and 7. From bay 0 the nearest is bay 7, from bay 7 it is bay 0, and from bay 4 it is bay 7.

Example 3

Input: codes = [4, 4, 4], asks = [0, 1, 2]
Output: [1, 1, 1]

With only three bays and one code, every bay has a neighbour holding the same code one step away.

Constraints

1 <= asks.length <= codes.length <= 10^5
1 <= codes[i] <= 10^6
0 <= asks[j] < codes.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 nearest_twin_steps(codes: list[int], asks: list[int]) -> list[int]:

Java

public List<Integer> nearestTwinSteps(int[] codes, int[] asks)

The intended approach

Time O(n + q) for n bays and q asksSpace O(n) for the buckets and the per-bay answers

Bucket the bay numbers by the code they hold. Inside one bucket the numbers come out in increasing order, and turning the carousel either way from a given bay reaches that bucket's ring-neighbours before any of its other entries, so only those two ever need measuring. Take the shorter of the two directions for each of the two neighbours, keep the smaller, and leave a bay alone when its bucket holds nothing else. Answers are then read straight off per ask.

Pythonentry point: nearest_twin_steps

def nearest_twin_steps(codes, asks):
    # Round the ring, the closest bay with a given code is one of that code's two
    # ring-neighbours in the sorted list of its positions.
    # time O(n + q), space O(n)
    n = len(codes)
    where = {}
    for i in range(n):
        where.setdefault(codes[i], []).append(i)
    answer = [-1] * n
    for spots in where.values():
        m = len(spots)
        if m < 2:
            continue
        for p in range(m):
            here = spots[p]
            back = spots[p - 1]
            fwd = spots[(p + 1) % m]
            a = abs(here - back)
            a = a if a < n - a else n - a
            b = abs(here - fwd)
            b = b if b < n - b else n - b
            answer[here] = a if a < b else b
    return [answer[q] for q in asks]

The obvious version

Handle every ask on its own: walk all n bays, and for each one holding the same code as the starting bay work out the shorter of the two ways round the ring, keeping the smallest. Report -1 when the walk finds nothing.

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_nearest_twin_steps(codes, asks):
    n = len(codes)
    out = []
    for q in asks:
        best = -1
        for j in range(n):
            if j == q or codes[j] != codes[q]:
                continue
            gap = abs(j - q)
            if n - gap < gap:
                gap = n - gap
            if best < 0 or gap < best:
                best = gap
        out.append(best)
    return out

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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