1061MediumLinked ListTwo Pointers

Swapping Two Cars in a Chain

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Trains the technique from

LeetCode 1721Swapping Nodes in a Linked 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 conveyor is built from cars clipped one behind another. Each car carries a load and a clip to the car behind it, and the car at the end has an empty clip. Nothing on the conveyor records how many cars it holds.

Because the harness passes plain JSON, the conveyor reaches you as chain, listing the loads in clip order from the front car.

Swap the loads of the k-th car counted from the front and the k-th car counted from the end, both counted from one. Return the conveyor afterwards.

Examples

Example 1

Input: chain = [9, 8, 7], k = 1
Output: [7, 8, 9]

The first car from the front and the first from the end are the two ends of the conveyor, so their loads change places.

Example 2

Input: chain = [4, 5], k = 2
Output: [5, 4]

With two cars, the second from the front is the last car and the second from the end is the first, so again the two ends swap.

Example 3

Input: chain = [1, 2, 3], k = 2
Output: [1, 2, 3]

With three cars, the second from the front and the second from the end are the same middle car, so nothing changes.

Constraints

1 <= k <= chain.length <= 10^5
0 <= chain[i] <= 100

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 swap_nodes(chain: list[int], k: int) -> list[int]:

Java

public int[] swapNodes(int[] chain, int k)

The intended approach

Time O(n) over the conveyor's length.Space O(1) beyond the conveyor itself.

The car counted `k` from the front is easy to reach by stepping forward `k - 1` times. The one counted `k` from the end is the trick, since the length is not recorded anywhere. Send a second marker forward from the front at the same time, but starting only once the first marker has taken its `k - 1` steps. The two markers then stay `k - 1` cars apart, so when the leading one reaches the last car, the trailing one stands exactly `k` cars from the end. Swap the two loads and the conveyor is done. One pass, and nothing held but two markers, which is what matters when only the clips can be followed. When the two markers land on the same car, the swap changes nothing, which needs no special case.

Pythonentry point: swap_nodes

def swap_nodes(chain, k):
    out = list(chain)
    front = k - 1
    behind = 0
    lead = front
    while lead < len(out) - 1:
        lead += 1
        behind += 1
    out[front], out[behind] = out[behind], out[front]
    return out

The obvious version

Count the cars first, then work out where the car counted from the end sits by subtracting, and swap the two loads. Needs the length up front, which is exactly what following clips alone cannot give you, but is straightforward once the whole conveyor is in hand.

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_swap_nodes(chain, k):
    out = list(chain)
    length = 0
    for _load in out:
        length += 1
    front = k - 1
    behind = length - k
    keep = out[front]
    out[front] = out[behind]
    out[behind] = keep
    return out

How this problem was checked

Reference solution executed against 15 test cases — 15 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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