0272MediumArrayTwo PointersGreedySortingTimsort

Gondola Cabins for the Ski Queue

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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 ski lift sends up cabins that seat at most two riders each, and a cabin may carry at most limit kilograms of rider weight in total. riders[i] is the weight in kilograms of the i-th person waiting, given in no particular order.

Everyone in the queue has to get up the mountain, and riders may be seated in any grouping you like as long as both cabin rules hold. Return the smallest number of cabins that will carry the whole queue.

No single rider ever exceeds limit, so the queue can always be cleared.

Examples

Example 1

Input: riders = [9, 2, 8, 3, 7, 4], limit = 11
Output: 3

Seating 2 with 9, 3 with 8 and 4 with 7 fills three cabins at 11 kg each, which seats all six riders inside both rules.

Example 2

Input: riders = [3, 4, 5], limit = 9
Output: 2

One cabin takes 3 and 5 for 8 kg and the next takes 4 on its own.

Example 3

Input: riders = [1, 1, 1, 1], limit = 4
Output: 2

A cabin seats two riders however light they are, so the four riders fill two cabins carrying 2 kg each.

Constraints

1 <= riders.length <= 5 * 10^4
1 <= riders[i] <= limit <= 3 * 10^4

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 gondola_cabin_count(riders: list[int], limit: int) -> int:

Java

public int gondolaCabinCount(int[] riders, int limit)

The intended approach

Time O(n log n)Space O(n) for the sorted copy

Sort the queue by weight and keep one index at the lightest rider left and one at the heaviest. Each step launches one cabin holding the heaviest rider left; the lightest rider left is seated beside them when their combined weight is within the limit, and otherwise stays in the queue. The heavy index always steps down, so every rider is placed after at most n steps.

Pythonentry point: gondola_cabin_count

def gondola_cabin_count(riders, limit):
    # Put the queue in weight order, then read it from both ends at once: the
    # heaviest rider left always boards the next cabin, and the lightest rider
    # left joins that cabin when the two of them fit inside the limit.
    # Time: O(n log n) for the sort plus one O(n) sweep.
    # Space: O(n) for the sorted copy the sort builds.
    order = sorted(riders)
    lo = 0
    hi = len(order) - 1
    cabins = 0
    while lo <= hi:
        if order[lo] + order[hi] <= limit:
            lo += 1
        hi -= 1
        cabins += 1
    return cabins

The obvious version

Subset dynamic programming: dp over bitmasks of seated riders, where each step seats the lowest-indexed remaining rider either alone or beside any one other remaining rider whose combined weight fits. dp on the full mask is the answer.

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_gondola_cabin_count(riders, limit):
    n = len(riders)
    size = 1 << n
    INF = n + 1
    dp = [INF] * size
    dp[0] = 0
    for mask in range(size):
        if dp[mask] == INF:
            continue
        i = 0
        while i < n and (mask >> i & 1):
            i += 1
        if i == n:
            continue
        alone = mask | (1 << i)
        if dp[mask] + 1 < dp[alone]:
            dp[alone] = dp[mask] + 1
        for j in range(i + 1, n):
            if mask >> j & 1:
                continue
            if riders[i] + riders[j] <= limit:
                pair = alone | (1 << j)
                if dp[mask] + 1 < dp[pair]:
                    dp[pair] = dp[mask] + 1
    return dp[size - 1]

How this problem was checked

Reference solution executed against 8 test cases — 8 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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