0711MediumTwo PointersString

Sliding Tokens in the Chute

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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 single-file chute holds sliding tokens. A configuration of the chute is written as a string, one character per position: '-' is an empty position, 'L' is a token that can only travel toward the front of the chute (lower positions) and 'R' is a token that can only travel toward the back (higher positions).

One move is either of:

pick an 'L' whose immediately preceding position is empty, and slide it one position forward into that empty position;
pick an 'R' whose immediately following position is empty, and slide it one position back into that empty position.

A token can never pass through another token, and a token never changes kind.

Given the current configuration chute and a wanted configuration target, both of the same length, return true when some sequence of moves turns chute into target, and false otherwise. Zero moves is allowed.

Examples

Example 1

Input: chute = "-LR-", target = "L--R"
Output: true

Sliding the `L` one position forward gives `L-R-`, then sliding the `R` one position back gives `L--R`. Both moves used an adjacent empty position in the token's own direction.

Example 2

Input: chute = "RL-", target = "-RL"
Output: false

The `R` sits at position 0 with the `L` immediately behind it, so it has no empty position to slide into, and the `L` has no empty position in front of it either. No move is possible at all, so `-RL` is out of reach.

Example 3

Input: chute = "RR-", target = "R--"
Output: false

A move never removes a token, so a configuration with two `R` tokens can never become one with a single `R`.

Constraints

1 <= chute.length <= 10^4
chute.length == target.length
chute and target each consist only of the characters '-', '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 chute_reachable(chute: str, target: str) -> bool:

Java

public boolean chuteReachable(String chute, String target)

The intended approach

Time O(n) for a chute of n positions: each pointer advances only forward.Space O(1).

A move never changes which tokens are present nor their order along the chute, since nothing may pass anything else, so the two configurations must list the same tokens in the same order once the empties are ignored. Beyond that, each token only has to be able to get from where it is to where it is wanted, which for an `'L'` means the wanted position is no further back and for an `'R'` means it is no further forward. Two pointers walk the configurations together, skipping empties, checking the pair of tokens match and that the position moved in the permitted direction; both pointers must run out at the same time.

Pythonentry point: chute_reachable

def chute_reachable(chute, target):
    # O(n), O(1) extra: walk both configurations with one pointer each, skipping
    # empties, and check the tokens line up and each one only has to travel the
    # way it is allowed to.
    n = len(chute)
    i = 0
    j = 0
    while True:
        while i < n and chute[i] == "-":
            i += 1
        while j < n and target[j] == "-":
            j += 1
        if i == n or j == n:
            return i == n and j == n
        if chute[i] != target[j]:
            return False
        if chute[i] == "L" and i < j:
            return False
        if chute[i] == "R" and i > j:
            return False
        i += 1
        j += 1

The obvious version

Breadth-first search over configurations, generating every legal single slide from each one and stopping when the wanted configuration turns up. Explores the whole reachable set, so it makes no claim about token order or travel direction, and is only usable on very short chutes.

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_chute_reachable(chute, target):
    if chute == target:
        return True
    n = len(chute)
    seen = {chute}
    frontier = [chute]
    while frontier:
        nxt = []
        for cur in frontier:
            for i in range(n):
                if cur[i] == "L" and i > 0 and cur[i - 1] == "-":
                    cand = cur[:i - 1] + "L-" + cur[i + 1:]
                elif cur[i] == "R" and i + 1 < n and cur[i + 1] == "-":
                    cand = cur[:i] + "-R" + cur[i + 2:]
                else:
                    continue
                if cand == target:
                    return True
                if cand not in seen:
                    seen.add(cand)
                    nxt.append(cand)
        frontier = nxt
    return False

How this problem was checked

Reference solution executed against 24 test cases — 24 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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