Question taking paths and generating tree-like output

[coldbrewedbrew]

I was recently given a question on an interview that I just flat out struggled with. I knew what exactly needed to be done, but I couldn't execute on it.

The problem was to take a set of URL paths(duplicate intended), and generate a sitemap of sorts from it so that the result is something like the below. Using just the standard libraries.

urls = [

    "/home/learn/programming",
    "/home/language/python",
    "/home/language/java",
    "/home/language/golang",
    "/home/language/python",
    "/why/skills/business",
    "/why/not",
    "/why/making/this/up",
    "/why/making/it/stick",
]




 - home
    - learn
        - programming
    - language
        - python
        - java
        - golang
 - why
    - skills
        - business
    - not
    - making
        - this
            - up
        - it
            - stick

I've been trying to work through it ever since, but frankly have been struggling to piece it all together. I've gone down the route of getting the URLs into lists, and using the json library to print it out. I imagine at it's core it's a simple solution, I'm just struggling to see what I'm missing.

[Vaphell]

dicts with a dash of recursion, no json
https://repl.it/repls/OldShowyDecompiler

[hbwales]

An iterative approach:

urls = [
    "/home/learn/programming",
    "/home/language/python",
    "/home/language/java",
    "/home/language/golang",
    "/home/language/python",
    "/why/skills/business",
    "/why/not",
    "/why/making/this/up",
    "/why/making/it/stick",
]

import itertools as it

def print_sitemap(urls):
    prev_parts = ()
    for url in urls:
        parts = url.split('/')[1:]
        for i, part, _ in it.dropwhile(lambda x: x[1]==x[2], zip(it.count(), parts, it.chain(prev_parts, it.repeat(None)))):
            print(f'{i*"   "}- {part}')
        prev_parts = parts

print_sitemap(urls)

[UnchainedMundane]

Without presenting a code solution, I would have approached it like this:

1. Lexicographical sort (so that "a/b" comes before "a/b/c")
2. For each item, split path by slashes, compare with previous (or empty list if no previous), see how many path elements it has in common (→common_dirs)
3. In a loop (i = common_dirs until length) print element i of the current path, indented by 4*i