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Railcar Chain Group Flips

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Trains the technique from

LeetCode 2074Reverse Nodes in Even Length Groups

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 holds one chain of coupled railcars. The chain reaches you as a list cars of car weights in order from the locomotive end to the tail, so cars[0] is the car coupled straight behind the locomotive and each following entry is the next car back.

Starting at the locomotive end, the chain is carved into consecutive groups. The first group takes 1 car, the second takes 2 cars, the third takes 3, and each group after that takes one more car than the group before it. The final group takes whatever cars are still left, which can be fewer than its turn calls for.

Every group holding an even number of cars has its cars put back in the opposite order, within the same stretch of the chain. Groups holding an odd number of cars are left as they are. Group boundaries never move, and no car ever leaves the group it landed in.

Return the weights of the chain in the resulting order. Do the rearranging by uncoupling and recoupling cars inside the chain rather than by building a fresh chain alongside it, and use only a constant amount of extra space beyond the chain itself.

Examples

Example 1

Input: cars = [40, 7, 19, 5, 26, 12, 33]
Output: [40, 19, 7, 5, 26, 12, 33]

The groups hold 1, 2, 3 and 1 cars, so only the pair 7 and 19 changes places; the group of three and the lone car at the tail stay as they were.

Example 2

Input: cars = [4, 9, 2, 6, 1]
Output: [4, 2, 9, 1, 6]

The groups hold 1, 2 and 2 cars, because only two cars are left for the third group; both pairs change places.

Example 3

Input: cars = [61, 14, 88, 3, 47, 20, 75, 9, 52, 36, 28, 66]
Output: [61, 88, 14, 3, 47, 20, 36, 52, 9, 75, 66, 28]

The groups hold 1, 2, 3, 4 and 2 cars; the groups of two and of four are put back in the opposite order and the groups of one and three are untouched.

Constraints

1 <= cars.length <= 10^5
0 <= cars[i] <= 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 flip_even_groups(cars: list[int]) -> list[int]:

Java

public int[] flipEvenGroups(int[] cars)

The intended approach

Time O(n)Space O(1) beyond the chain handed in

Track the index where the current group begins and the size its turn calls for, and clamp the end of the group to the length of the chain so the final group reports the number of cars it really holds. When that count is even, reverse the stretch in place with an index from each end walking towards the middle. Then advance the start index to the group end and raise the size by one.

Pythonentry point: flip_even_groups

def flip_even_groups(cars):
    # Walk the group boundaries with a running start index and a group size that
    # grows by one each time, clamping the last group to what is left. When the
    # clamped size is even, swap the group's ends inwards in place.
    # time O(n), space O(1) beyond the chain handed in
    n = len(cars)
    start = 0
    size = 1
    while start < n:
        end = start + size
        if end > n:
            end = n
        if (end - start) % 2 == 0:
            left = start
            right = end - 1
            while left < right:
                cars[left], cars[right] = cars[right], cars[left]
                left += 1
                right -= 1
        start = end
        size += 1
    return cars

The obvious version

Slice the chain into groups of growing size, reverse the slice with a stride of minus one whenever the slice length is even, and glue the slices onto a fresh output list.

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_flip_even_groups(cars):
    out = []
    at = 0
    size = 1
    while at < len(cars):
        chunk = cars[at:at + size]
        if len(chunk) % 2 == 0:
            chunk = chunk[::-1]
        out.extend(chunk)
        at += size
        size += 1
    return out

How this problem was checked

Reference solution executed against 11 test cases — 11 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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