1132MediumStringStackDepth-First Search

The Longest Route Down the Catalogue

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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 catalogue arrives as one string listing. Entries are separated by the newline character, and each entry is preceded by as many tab characters as its depth, so an entry at depth 0 has none, one at depth 1 has one, and so on. An entry at depth d + 1 sits inside the nearest earlier entry at depth d.

An entry whose name holds a dot is a crate, and every other entry is a shelf.

The route to an entry is the names from the outermost shelf down to the entry itself, joined by single / characters. Return the length of the longest route to a crate, or 0 when the catalogue holds no crate at all.

Examples

Example 1

Input: listing = "dir\n\tsub\n\tfile.a\n\tsub2\n\t\tfile.bb"
Output: 16

The deeper crate's route joins three names with two slashes for sixteen characters, while the crate one level in gives only ten.

Example 2

Input: listing = "shelf"
Output: 0

The one entry holds no dot, so it is a shelf and the catalogue holds no crate at all.

Example 3

Input: listing = "a.b"
Output: 3

The single entry holds a dot, so it is a crate and its route is just its own name.

Constraints

1 <= listing.length <= 10^4
listing may hold English letters, newline characters, tab characters, dots, spaces and digits
every name holds at least one character

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 length_longest_path(listing: str) -> int:

Java

public int lengthLongestPath(String listing)

The intended approach

Time O(n) over the listing.Space O(d) for one prefix length per depth.

The route to an entry is its own name added to the route of the shelf it sits inside, so the only thing worth carrying along is one number per depth: the length of the route prefix that entries at that depth inherit, counting the trailing slash. Walk the entries in the order they are listed. Count the leading tabs to get the depth and take the rest as the name. A shelf at depth `d` sets the prefix for depth `d + 1` to its own inherited prefix plus its name plus one for the slash. A crate at depth `d` finishes a route, so its length is the inherited prefix plus its own name, with no trailing slash, and the largest such length is the answer. Overwriting the prefix for a depth as each new shelf appears is enough, because entries always follow the shelf they sit inside and no later entry at that depth can belong to an earlier shelf. Nothing needs unwinding when the depth drops.

Pythonentry point: length_longest_path

def length_longest_path(listing):
    best = 0
    prefix = {0: 0}
    for entry in listing.split("\n"):
        depth = 0
        while depth < len(entry) and entry[depth] == "\t":
            depth += 1
        name = entry[depth:]
        if "." in name:
            span = prefix[depth] + len(name)
            if span > best:
                best = span
        else:
            prefix[depth + 1] = prefix[depth] + len(name) + 1
    return best

The obvious version

Keep the actual trail of names from the outermost shelf downwards. For each entry, cut the trail back to its depth, push its name on, and when it is a crate build the whole route by joining the trail with slashes and measure it. Costs building a fresh string for every crate.

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_length_longest_path(listing):
    best = 0
    trail = []
    for entry in listing.split("\n"):
        depth = 0
        while depth < len(entry) and entry[depth] == "\t":
            depth += 1
        name = entry[depth:]
        while len(trail) > depth:
            trail.pop()
        trail.append(name)
        if "." in name:
            route = "/".join(trail)
            if len(route) > best:
                best = len(route)
    return best

How this problem was checked

Reference solution executed against 15 test cases — 15 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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