0324EasyStringSimulation

Shuttle Back On The Dock

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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 floor shuttle in a parts warehouse begins parked on its charging dock and then plays back a tape of single-letter steps, moves, one step per letter, each step covering one tile:

'U' moves it one tile up the aisle,
'D' moves it one tile down the aisle,
'L' moves it one tile to the left,
'R' moves it one tile to the right.

The floor is unobstructed, so every step is carried out exactly as written and the shuttle may pass over the dock tile as often as it likes mid-tape.

Return true when the shuttle is sitting on the dock tile once the whole tape has been played back, and false when it stops anywhere else.

Examples

Example 1

Input: moves = "RRLL"
Output: true

Two steps right then two steps left leave the shuttle on the dock tile, so the answer is true.

Example 2

Input: moves = "RRUL"
Output: false

The shuttle finishes one tile to the right of the dock and one tile up the aisle from it, so it is not parked.

Example 3

Input: moves = "ULDR"
Output: true

The shuttle walks a small loop, passing three other tiles, and its last step puts it back on the dock.

Example 4

Input: moves = "D"
Output: false

A single step down the aisle leaves the shuttle one tile away from the dock.

Constraints

1 <= moves.length <= 2 * 10^4
moves only contains the characters 'U', 'D', 'L' and 'R'.

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 ends_on_dock(tape: str) -> bool:

Java

public boolean endsOnDock(String tape)

The intended approach

Time O(n)Space O(1)

Keep one signed offset for the aisle direction and one for the sideways direction, adjust the matching offset per letter, and report whether both offsets are zero at the end. Nothing about the intermediate path needs to be kept.

Pythonentry point: ends_on_dock

def ends_on_dock(tape):
    # One pass, two running offsets, nothing stored per step.
    # Time O(n). Space O(1).
    across = 0
    along = 0
    for ch in tape:
        if ch == "U":
            along += 1
        elif ch == "D":
            along -= 1
        elif ch == "L":
            across -= 1
        else:
            across += 1
    return across == 0 and along == 0

The obvious version

Four separate full scans with the built-in character count, then compare the opposing letters pairwise.

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_ends_on_dock(tape):
    return tape.count("U") == tape.count("D") and tape.count("L") == tape.count("R")

How this problem was checked

Reference solution executed against 9 test cases — 9 passed
4 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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