0782EasyArray

Shortest Run Round The Monorail Loop

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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.

An airport monorail serves terminals numbered 0 to n - 1 set out on a closed loop. hop[i] is the length of track joining terminal i to terminal i + 1 for every i < n - 1, and the last entry hop[n - 1] is the length of track joining terminal n - 1 back to terminal 0, which closes the loop.

A shuttle may set off either way round the loop and must stay in that direction until it arrives. Given the terminal a passenger boards at and the terminal they get off at, return the smallest total track length the shuttle can cover to carry them there.

Boarding and alighting at the same terminal covers no track at all.

Examples

Example 1

Input: hop = [4,7,2,9,5], boarding = 1, alighting = 4
Output: 9

Running from terminal 1 down through terminal 0 covers hop[0] = 4 and then hop[4] = 5, so that direction totals 9. Running the other way covers hop[1] = 7, hop[2] = 2 and hop[3] = 9, a total of 18.

Example 2

Input: hop = [1,20,20,20], boarding = 1, alighting = 0
Output: 1

Terminals 0 and 1 are joined directly by hop[0] = 1, and the way round through terminals 2 and 3 covers 20 + 20 + 20 = 60.

Example 3

Input: hop = [6,6,6,6], boarding = 3, alighting = 3
Output: 0

The passenger gets off where they got on, so the shuttle covers no track.

Constraints

1 <= hop.length <= 10^4
0 <= hop[i] <= 10^4
0 <= boarding <= 10^4 - 1
0 <= alighting <= 10^4 - 1
boarding and alighting are terminal numbers, so both are below hop.length.

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 shortest_loop_run(hop: list[int], boarding: int, alighting: int) -> int:

Java

public int shortestLoopRun(int[] hop, int boarding, int alighting)

The intended approach

Time O(n) for the single pass over hop.Space O(1) beyond the input array.

Order the two terminals so the lower number comes first. One pass over `hop` adds up the whole loop while separately adding up the segments whose index sits in the half-open range between the two terminals, which is the arc running through increasing terminal numbers. The remaining arc is the loop total minus that, and the answer is whichever of the two is smaller.

Pythonentry point: shortest_loop_run

def shortest_loop_run(hop, boarding, alighting):
    # The loop splits into exactly two arcs at the two terminals. Sum the arc that
    # runs through increasing terminal numbers, and the other arc is the whole loop
    # minus it. Time O(n) for one pass over hop, space O(1) beyond the input.
    lo = boarding if boarding < alighting else alighting
    hi = alighting if boarding < alighting else boarding
    total = 0
    arc = 0
    for i, length in enumerate(hop):
        total += length
        if lo <= i < hi:
            arc += length
    other = total - arc
    return arc if arc < other else other

The obvious version

Simulate the shuttle. Step from terminal to terminal in the increasing direction, wrapping past the last terminal, adding each segment length until the destination is reached; repeat stepping in the decreasing direction; return the smaller running total.

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_shortest_loop_run(hop, boarding, alighting):
    n = len(hop)
    here = boarding
    forward = 0
    while here != alighting:
        forward += hop[here]
        here = (here + 1) % n
    here = boarding
    backward = 0
    while here != alighting:
        prev = (here - 1) % n
        backward += hop[prev]
        here = prev
    return min(forward, backward)

How this problem was checked

Reference solution executed against 12 test cases — 12 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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