0351MediumStringStack

Canonical Store Locator

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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 artifact store addresses every folder with a locator that begins at the root, written as /. A locator is a run of pieces separated by /, and the store reads it like this:

a piece of . means stay where you are;
a piece of .. means step back to the folder that contains the current one, and at the root it does nothing at all;
any other piece is a folder name, and names may contain letters, digits, underscores and even runs of dots such as ..., which are ordinary names and not steps;
separators may be doubled up, and // behaves exactly like /.

The canonical form of a locator starts with /, puts exactly one / between consecutive folder names, and never ends with / unless it is the root by itself. Given path, return its canonical form.

Examples

Example 1

Input: path = "/store//alpha/"
Output: "/store/alpha"

The doubled separator collapses to one and the trailing separator is dropped, leaving the two folder names in place.

Example 2

Input: path = "/a/./b/../../c/"
Output: "/c"

`.` leaves the position alone, the first `..` undoes `b` and the second undoes `a`, so only `c` is left below the root.

Example 3

Input: path = "/_x/y/.._/.."
Output: "/_x/y"

`.._` is an ordinary folder name, so it is entered like any other, and the final `..` steps back out of it.

Example 4

Input: path = "/..."
Output: "/..."

A piece of three dots is a folder name, so the canonical form keeps it.

Constraints

1 <= path.length <= 3000
path holds English letters, digits, '.', '/' and '_' only.
path is a valid absolute locator, so it starts with '/'.

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 simplify_path(path: str) -> str:

Java

public String simplifyPath(String path)

The intended approach

Time O(len(path))Space O(len(path))

Cut the locator at every separator, which turns doubled separators into empty pieces that can be skipped along with `.`. A `..` piece pops the folder currently on top of the stack, or is dropped when the stack is already at the root. Joining the surviving names with single separators after a leading one gives the canonical form, and an empty stack yields the root.

Pythonentry point: simplify_path

def simplify_path(path):
    # Split the locator at the separators and push names onto a stack.  ".." pops the
    # name on top when there is one and is ignored at the root, "." and empty pieces
    # are noise.  Rebuilding from the stack puts back exactly one separator between
    # names and none at the end.
    # Time O(len(path)).  Space O(len(path)) for the pieces and the stack.
    stack = []
    for piece in path.split("/"):
        if piece == "" or piece == ".":
            continue
        if piece == "..":
            if stack:
                stack.pop()
            continue
        stack.append(piece)
    return "/" + "/".join(stack)

The obvious version

Collect the pieces character by character without any split, drop the `.` pieces, then keep finding the leftmost `..` and delete it together with the piece before it until none is left. Rebuild the locator from what remains.

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_canonical_locator(path):
    pieces = []
    current = ""
    for ch in path:
        if ch == "/":
            if current != "":
                pieces.append(current)
                current = ""
        else:
            current += ch
    if current != "":
        pieces.append(current)
    pieces = [p for p in pieces if p != "."]
    while True:
        hop = -1
        for i in range(len(pieces)):
            if pieces[i] == "..":
                hop = i
                break
        if hop < 0:
            break
        if hop == 0:
            pieces = pieces[1:]
        else:
            pieces = pieces[:hop - 1] + pieces[hop + 1:]
    out = ""
    for p in pieces:
        out += "/" + p
    if out == "":
        out = "/"
    return out

How this problem was checked

Reference solution executed against 11 test cases — 11 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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