re creative solution for Words Order by juestr - python coding challenges

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re solution in Creative category for Words Order by juestr

import re

def words_order(text, words):
    return len(set(words)) == len(words) and \
        not not re.search(r'\b' + r'\b.*\b'.join(words) + r'\b', text)

May 23, 2020

Comments:

koborC on Nov. 2, 2021, 8:37 a.m.
Do the "not not ..." means bool(...)?

juestr on Nov. 2, 2021, 11:50 a.m.
Yes. Don't do this in clean code.

jakubbortlik on Nov. 15, 2022, 2:44 p.m.
I wrote a very similar solution (except I used `bool()` :) and a non-greedy `r'\b.*?\b'`), but your solution just like mine does not satisfy the condition that the first occurrence of a word in `text` should be considered. So this fails for `assert words_order("here world here here", ["world", "here"]) == False`, but the checks are not complete and do non contain these kind of inputs, and so the solution passes.

juestr on Nov. 15, 2022, 3:06 p.m.
Yeah, but it depends somewhat on the interpretation of "any word can appear more than once in a text - use only the first one". Which first? From the left? Or the first found in a correct position and then simply ignore extra occurrences? In that case your example should be `True`. I chose the more permissive option. ;-)

H0r4c3 on April 29, 2023, 10:31 a.m.
Nice solution! Regex for the win!

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