0103EasyArrayHash TableSliding Window

Nearby Repeat Trim

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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 lathe operator keeps a trim log. trims[i] is the signed correction, in micrometres, that was dialled in before pass i: a negative figure pulls the cut back, a positive figure pushes it forward, and 0 means the pass ran with the dial untouched.

Quality control cares about corrections that come back around quickly, because a value reused soon after its last use points at a machine that is drifting rather than settling. You are given the log and an integer span.

Report whether some correction figure was dialled in on two separate passes whose pass numbers are no more than span apart. Return true when such a pair of passes exists and false when none does. A pass is never a repeat of itself, so a span of 0 can only ever give false.

Examples

Example 1

Input: trims = [4, -2, 7, -2, 9], span = 2
Output: true

The correction -2 was dialled in before pass 1 and again before pass 3, and those passes sit 2 apart, which the span allows.

Example 2

Input: trims = [5, -3, 5], span = 1
Output: false

Only the figure 5 was reused, on passes 0 and 2. Those passes are 2 apart, so with a span of 1 the log counts as settled.

Example 3

Input: trims = [8, -1, 3, 8], span = 3
Output: true

The reused figure 8 sits on the first and last passes, exactly 3 apart, so the final pass still has to be weighed against the whole span behind it.

Constraints

1 <= trims.length <= 10^5
-10^9 <= trims[i] <= 10^9
0 <= span <= 10^5

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 has_nearby_repeat(trims: list[int], span: int) -> bool:

Java

public boolean hasNearbyRepeat(int[] trims, int span)

The intended approach

Time O(n)Space O(min(n, span))

Sweep the log once while holding the `span` most recent figures in a set. Before reading pass `i`, evict the figure from pass `i - span - 1`, which has just fallen out of reach, so a hit in the set is guaranteed to be close enough. Membership and eviction are both constant time, so one pass over the log settles it.

Pythonentry point: has_nearby_repeat

def has_nearby_repeat(trims, span):
    if span <= 0:
        return False
    recent = set()
    for i, value in enumerate(trims):
        if i > span:
            recent.discard(trims[i - span - 1])
        if value in recent:
            return True
        recent.add(value)
    return False

The obvious version

For every pass, compare it against each of the next `span` passes directly, O(n * span) 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_nearby_repeat(trims, span):
    n = len(trims)
    for i in range(n):
        for j in range(i + 1, min(n, i + span + 1)):
            if trims[i] == trims[j]:
                return True
    return False

How this problem was checked

Reference solution executed against 12 test cases — 12 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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