0141MediumLinked ListTwo Pointers

Unhook the Nth Wagon from the Tail

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Trains the technique from

LeetCode 19Remove Nth Node From End of List

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 shunting yard models a train as a one-way coupling chain: every wagon carries a load reading and points at the wagon coupled behind it, and the rear wagon points at nothing.

The harness passes plain JSON, so the train arrives as the array wagons, holding the load readings from the front wagon back to the rear one. Your answer takes the same shape: the readings that remain, front wagon first, or an empty array when nothing is left.

Counting from the rear, where the rear wagon is number 1, unhook wagon number n and couple its two neighbours to each other. The crew has one walk down the train from the front, so rebuild the chain, then find and unhook the wagon in a single pass over the couplings while holding only a fixed number of wagon references. Measuring the train's length first and walking it again does not count as a single pass.

Examples

Example 1

Input: wagons = [12, 5, 30, 8, 21], n = 2
Output: [12, 5, 30, 21]

From the rear, the wagon loaded with 21 is number 1 and the one loaded with 8 is number 2, so 8 rolls away and 30 couples straight to 21.

Example 2

Input: wagons = [4, 7], n = 2
Output: [7]

Wagon 2 from the rear is the front wagon here, so the train now begins at the wagon loaded with 7.

Example 3

Input: wagons = [9], n = 1
Output: []

A single wagon is also the rear wagon, and unhooking it leaves an empty train.

Constraints

1 <= wagons.length <= 30
0 <= wagons[i] <= 100
1 <= n <= wagons.length
Only a fixed number of wagon references may be held, and the couplings may be walked once

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 unhook_from_tail(wagons: list[int], n: int) -> list[int]:

Java

public int[] unhookFromTail(int[] wagons, int n)

The intended approach

Time O(n)Space O(n) for the chain rebuilt from the input, plus O(1) wagon references beyond it

Rebuild the chain behind a stand-in engine coupling so the front wagon can be unhooked the same way as any other. Send a lead reference n couplings ahead, then advance the lead and a trailing reference together until the lead sits on the rear wagon; the trailing reference is now parked exactly one coupling ahead of the wagon to remove, so redirecting its coupling past that wagon closes the gap in one walk.

Pythonentry point: unhook_from_tail

def unhook_from_tail(wagons, n):
    class Wagon:
        __slots__ = ("load", "nxt")

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

    engine = Wagon(0)
    tail = engine
    for load in wagons:
        tail.nxt = Wagon(load)
        tail = tail.nxt
    lead = engine
    for _ in range(n):
        lead = lead.nxt
    trail = engine
    while lead.nxt is not None:
        lead = lead.nxt
        trail = trail.nxt
    trail.nxt = trail.nxt.nxt
    out = []
    node = engine.nxt
    while node is not None:
        out.append(node.load)
        node = node.nxt
    return out

The obvious version

Take the length, turn the count from the rear into a position from the front, and copy every other reading across. Two passes and plain index arithmetic instead of a coupling walk.

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_unhook_from_tail(wagons, n):
    cut = len(wagons) - n
    kept = []
    for i in range(len(wagons)):
        if i != cut:
            kept.append(wagons[i])
    return kept

How this problem was checked

Reference solution executed against 9 test cases — 9 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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