0034MediumArrayMathTwo Pointers

Delay the Sample Buffer

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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 firmware ring buffer holds one signed sensor delta per slot in the integer array readings. Slot 0 is the oldest slot and the buffer wraps, so the slot after the last one is slot 0 again.

A technician wants the whole trace pushed later by delay slots: whatever sits in a slot ends up delay slots further along, and anything that runs off the end reappears at the front. The value of delay can be 0, and it can also be larger than the number of slots, in which case the trace simply wraps more than once.

Rearrange readings itself, using no more than a constant amount of extra space on top of the buffer, then return readings.

Examples

Example 1

Input: readings = [5, -2, 9, 4], delay = 1
Output: [4, 5, -2, 9]

Every delta moves one slot later and the 4 that fell off the end lands in slot 0.

Example 2

Input: readings = [7, 8, -3], delay = 5
Output: [8, -3, 7]

Five pushes across three slots is one full wrap plus two more, so the outcome matches a push of two.

Example 3

Input: readings = [6, 1, 4], delay = 0
Output: [6, 1, 4]

Nothing is pushed, so the buffer comes back exactly as it arrived.

Constraints

1 <= readings.length <= 10^5
-2^31 <= readings[i] <= 2^31 - 1
0 <= delay <= 10^5
Only a constant amount of extra space may be used

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 delay_buffer(readings: list[int], delay: int) -> list[int]:

Java

public int[] delayBuffer(int[] readings, int delay)

The intended approach

Time O(n)Space O(1)

Only `delay % n` matters, since a full wrap is a no-op. Reverse the whole buffer, which puts the tail block in front but backwards, then reverse the first `shift` slots and the remaining slots separately to restore the order inside each block. All three reversals swap pairs in place, so nothing beyond a couple of index variables is allocated.

Pythonentry point: delay_buffer

def delay_buffer(readings, delay):
    n = len(readings)
    shift = delay % n

    def flip(lo, hi):
        while lo < hi:
            readings[lo], readings[hi] = readings[hi], readings[lo]
            lo += 1
            hi -= 1

    flip(0, n - 1)
    flip(0, shift - 1)
    flip(shift, n - 1)
    return readings

The obvious version

Work on a copy and push the trace one slot at a time, `delay % n` times, lifting the last entry and dropping it at the front. Obviously correct, O(n * delay) in the worst case.

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_delay_buffer(readings, delay):
    data = list(readings)
    n = len(data)
    for _ in range(delay % n):
        tail = data.pop()
        data.insert(0, tail)
    return data

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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