0151MediumLinked ListTwo Pointers

Rundown Cue Shift

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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 radio show keeps its rundown as a one-way chain of cue cards. Each card carries a signed timing offset in seconds and a link to the card that comes after it, and the final card links to nothing.

Because the harness passes plain JSON, the chain reaches you as the array cues, listing the offsets from the leading card through to the final one. An empty array means there is no rundown at all. Your answer must come back in the same shape: the offsets of the rearranged chain, leading card first.

The producer wants the rundown wound forward by shift places. One place forward means the final card becomes the leader and everything else slides back one spot; shift places forward means the last shift cards move ahead of the rest as a block, with their own order among themselves untouched. The value of shift can run far past the number of cards, in which case the winding simply comes around the chain again as many times as it needs to. A rundown of no cards, or of one card, comes back exactly as it arrived.

Do the work the chain calls for: hold the rundown as linked cards and move cards by changing which card each one links to. Do not build the answer by sorting the offsets or by pasting slices of the array together.

Examples

Example 1

Input: cues = [6, -2, 9, 4, -7], shift = 2
Output: [4, -7, 6, -2, 9]

Winding forward twice moves the final two cards, holding 4 and -7 in that order, ahead of the other three.

Example 2

Input: cues = [3, 8, -5], shift = 7
Output: [-5, 3, 8]

Seven places around a chain of three cards is two full laps plus one more place, so only the final card ends up in front.

Example 3

Input: cues = [], shift = 4
Output: []

There are no cards to wind, so the rundown stays empty however large the shift is.

Constraints

0 <= cues.length <= 500
-100 <= cues[i] <= 100
0 <= shift <= 2 * 10^9

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 rundown_cue_shift(cues: list[int], shift: int) -> list[int]:

Java

public int[] rundownCueShift(int[] cues, int shift)

The intended approach

Time O(n)Space O(n) for the rebuilt chain and the returned offsets, with only a fixed number of card references beyond that

Winding forward by a whole number of laps changes nothing, so the first move is to reduce `shift` against the number of cards; that alone keeps a huge shift cheap. Rebuild the chain, join the final card back onto the leader so it becomes a ring, then step forward to the card that has to end up last and cut the ring open just after it. The card on the far side of that cut is the new leader, and no card is ever copied or reordered by hand.

Pythonentry point: rundown_cue_shift

def rundown_cue_shift(cues, shift):
    # Time: O(n) -- one pass to build and measure, one partial pass to find the cut.
    # Space: O(n) for the rebuilt cards and the offsets returned, O(1) extra references.
    class Card:
        __slots__ = ("offset", "nxt")

        def __init__(self, offset):
            self.offset = offset
            self.nxt = None

    lead = None
    final = None
    count = 0
    for offset in cues:
        card = Card(offset)
        if lead is None:
            lead = card
        else:
            final.nxt = card
        final = card
        count += 1
    if count == 0:
        return []
    places = shift % count
    if places:
        final.nxt = lead
        steps = count - places - 1
        cut = lead
        while steps:
            cut = cut.nxt
            steps -= 1
        lead = cut.nxt
        cut.nxt = None
    out = []
    card = lead
    while card is not None:
        out.append(card.offset)
        card = card.nxt
    return out

The obvious version

Wind the rundown forward one single place at a time, rebuilding the whole order on each place, and repeat that as many times as the reduced shift asks for.

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_rundown_cue_shift(cues, shift):
    order = list(cues)
    count = len(order)
    if count == 0:
        return []
    for _ in range(shift % count):
        moved = [order[count - 1]]
        for k in range(count - 1):
            moved.append(order[k])
        order = moved
    return order

How this problem was checked

Reference solution executed against 14 test cases — 14 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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