0143EasyArrayMath

Wheel Counter Tick

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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 turnstile keeps its running count on a row of mechanical drums, one digit per drum. The array digits reports what the drums currently show, with digits[0] sitting at the far left of the row.

One more person walks through, so the counter ticks up by a single step. A drum showing nine wraps back to zero and nudges the drum to its left; when the leftmost drum wraps, the housing exposes a fresh drum in front of it that now shows one. Give back the drum readings after the tick.

Examples

Example 1

Input: digits = [3, 1, 7]
Output: [3, 1, 8]

The rightmost drum has room to climb, so nothing else on the row moves.

Example 2

Input: digits = [6, 9]
Output: [7, 0]

The rightmost drum wraps to zero and nudges its neighbour from six to seven.

Example 3

Input: digits = [9, 9, 9]
Output: [1, 0, 0, 0]

Every drum wraps, so the housing exposes a new drum in front showing one.

Constraints

1 <= digits.length <= 100
0 <= digits[i] <= 9
The reading has no zero drum in front of its first significant drum.

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 plus_one(digits: list[int]) -> list[int]:

Java

public int[] plusOne(int[] digits)

The intended approach

Time O(n)Space O(1)

Walk the row from the right. The first drum below nine absorbs the tick and the walk stops there. Every nine passed on the way is set to zero, and if the walk runs off the left edge the whole row was nines, so a leading one is placed in front.

Pythonentry point: plus_one

def plus_one(digits):
    i = len(digits) - 1
    while i >= 0:
        if digits[i] < 9:
            digits[i] += 1
            return digits
        digits[i] = 0
        i -= 1
    return [1] + digits

The obvious version

Read the whole row as one integer, add one to it, then split the sum back into drum readings. No carry handling of its own.

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_tick_counter(digits):
    total = 0
    for d in digits:
        total = total * 10 + d
    total += 1
    out = []
    while total > 0:
        out.append(total % 10)
        total //= 10
    if not out:
        out.append(0)
    out.reverse()
    return out

How this problem was checked

Reference solution executed against 10 test cases — 10 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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