0662EasyArraySliding Window

Rewriting the Dial 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 mechanical dial carries n numbered slots arranged in a circle; ring[i] is the number engraved on slot i, and slot n - 1 sits immediately before slot 0 going clockwise. A service manual asks you to re-engrave the whole dial at once, using a single setting shift.

For every slot i, its new number is:

the sum of the numbers on the next shift slots clockwise, when shift > 0;
the sum of the numbers on the previous -shift slots anticlockwise, when shift < 0;
0, when shift == 0.

The counting wraps round the circle, and a slot never counts itself. Every new number is worked out from the numbers engraved before the rewrite, not from numbers written during it.

Return the list of new numbers, slot 0 first.

Examples

Example 1

Input: ring = [5, 7, 1, 4], shift = 2
Output: [8, 5, 9, 12]

Slot 0 reads slots 1 and 2, giving 7 + 1 = 8. Slot 1 reads slots 2 and 3, giving 1 + 4 = 5. Slot 2 wraps to slots 3 and 0, giving 4 + 5 = 9, and slot 3 wraps to slots 0 and 1, giving 5 + 7 = 12.

Example 2

Input: ring = [5, 7, 1, 4], shift = -1
Output: [4, 5, 7, 1]

Each slot takes the single number behind it, so slot 0 wraps back to slot 3 and reads 4, while slots 1, 2 and 3 read 5, 7 and 1.

Example 3

Input: ring = [3, 9, 2], shift = 0
Output: [0, 0, 0]

A setting of zero engraves 0 on every slot.

Constraints

1 <= ring.length <= 100
1 <= ring[i] <= 100
-(n - 1) <= shift <= n - 1, where n is the number of slots.

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 rewrite_ring(ring: list[int], shift: int) -> list[int]:

Java

public int[] rewriteRing(int[] ring, int shift)

The intended approach

Time O(n + |shift|): one pass to build the first run, then two additions per slot.Space O(1) beyond the returned list.

Every slot reads a run of `|shift|` consecutive slots, and consecutive slots read runs that overlap in all but one place. So work out the run for slot `0` by adding its `|shift|` numbers, then walk forward: each step subtracts the number that fell off the back of the run and adds the one that joined the front, with both indices taken modulo `n`. The starting offset is `1` for a clockwise setting and `n - |shift|` for an anticlockwise one, and a setting of zero is answered with all zeros.

Pythonentry point: rewrite_ring

def rewrite_ring(ring, shift):
    # One window of |shift| slots slides once round the dial; each step drops the
    # slot that left and adds the slot that entered. O(n) time.
    n = len(ring)
    out = [0] * n
    if shift == 0:
        return out
    width = shift if shift > 0 else -shift
    start = 1 if shift > 0 else n - width
    total = 0
    for t in range(width):
        total += ring[(start + t) % n]
    out[0] = total
    for i in range(1, n):
        total -= ring[(start + i - 1) % n]
        total += ring[(start + i - 1 + width) % n]
        out[i] = total
    return out

The obvious version

Take each slot in turn and walk the required number of steps round the circle one slot at a time, taking indices modulo `n`, adding up what it finds.

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_rewrite_ring(ring, shift):
    n = len(ring)
    out = []
    for i in range(n):
        acc = 0
        if shift > 0:
            for step in range(1, shift + 1):
                acc += ring[(i + step) % n]
        elif shift < 0:
            for step in range(1, -shift + 1):
                acc += ring[(i - step) % n]
        out.append(acc)
    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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