0229MediumLinked ListTwo Pointers

Retether the Sensor Pods

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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 survey tether carries a single file of sensor pods, each pod clipped to the one before it. The tether reaches you as pods, the list of calibration offsets in hanging order with the anchor end first, so pods[0] is the pod clipped to the anchor. Offsets may be negative. You are also given an integer threshold.

Re-clip the file so that every pod whose offset is below threshold ends up closer to the anchor than every pod whose offset is threshold or more. Inside each of those two stretches the pods keep the order they hung in when the tether arrived.

Work by unclipping whole pods and clipping them back on: an offset is never edited, and pods are not put in order by offset value. Return the offsets of the re-clipped tether in hanging order, anchor end first. The tether may arrive with no pods on it, in which case the answer is an empty list.

Examples

Example 1

Input: pods = [3, 9, 1, 7, 2], threshold = 5
Output: [3, 1, 2, 9, 7]

Offsets 3, 1 and 2 are below 5, and they reach the anchor end still in that order. Offsets 9 and 7 are 5 or more and follow behind them, also in the order they arrived.

Example 2

Input: pods = [6, 3, 8, 3, 9, 1], threshold = 4
Output: [3, 3, 1, 6, 8, 9]

The below-threshold pods 3, 3 and 1 come first in their arrival order, then 6, 8 and 9 in theirs. Two pods share the offset 3 and both are kept.

Example 3

Input: pods = [], threshold = 0
Output: []

The tether arrived with no pods clipped to it, so there is nothing to re-clip.

Constraints

0 <= pods.length <= 200
-100 <= pods[i] <= 100
-200 <= threshold <= 200

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 retether_pods(pods: list[int], threshold: int) -> list[int]:

Java

public int[] retetherPods(int[] pods, int threshold)

The intended approach

Time O(n)Space O(n) for the returned tether, O(1) beyond it

Walk the tether once, holding the tail of two separate stretches. Each pod is appended to the near stretch when its offset is below the threshold and to the far stretch otherwise, and appending rather than inserting is what keeps both stretches in arrival order. Clipping the far stretch onto the end of the near one produces the answer.

Pythonentry point: retether_pods

def retether_pods(pods, threshold):
    near = []
    far = []
    for offset in pods:
        if offset < threshold:
            near.append(offset)
        else:
            far.append(offset)
    near.extend(far)
    return near

The obvious version

Repeatedly sweep the tether and swap any neighbouring pair where a threshold-or-more pod sits ahead of a below-threshold pod, until n sweeps have run. Adjacent swaps never disturb the order inside either stretch. O(n^2).

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_retether_pods(pods, threshold):
    chain = list(pods)
    n = len(chain)
    for _ in range(n):
        for i in range(n - 1):
            if chain[i] >= threshold and chain[i + 1] < threshold:
                chain[i], chain[i + 1] = chain[i + 1], chain[i]
    return chain

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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