First clear solution for Words Order by book1978 - python coding challenges

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First solution in Clear category for Words Order by book1978

def words_order(text: str, words: list) -> bool:
    text = text.split()
    s = [text.index(i) if i in text else -1 for i in words]
    return s == sorted(s) and len(words) == len(set(words)) and -1 not in s


print("Example:")
print(words_order("hi world im here", ["world", "here"]))

assert words_order("hi world im here", ["world", "here"]) == True
assert words_order("hi world im here", ["here", "world"]) == False
assert words_order("hi world im here", ["world"]) == True
assert words_order("hi world im here", ["world", "here", "hi"]) == False
assert words_order("hi world im here", ["world", "im", "here"]) == True
assert words_order("hi world im here", ["world", "hi", "here"]) == False
assert words_order("hi world im here", ["world", "world"]) == False
assert words_order("hi world im here", ["country", "world"]) == False
assert words_order("hi world im here", ["wo", "rld"]) == False
assert words_order("", ["world", "here"]) == False
assert words_order("hi world world im here", ["world", "world"]) == False
assert (
    words_order("hi world world im here hi world world im here", ["world", "here"])
    == True
)

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Jan. 12, 2023

Comments:

Germ_2c646239b437492980bfd9ddf on April 6, 2023, 3:37 p.m.
Really nice solution. But maybe you could modify it so that, when a word isn't found, it just returns false instead of adding a -1 to the list? Sure, the code looks cleaner with a single return, but you are just making the program run for longer when you already determined that the result is false. The same goes for the check about the duplicates in words. Since you are doing it anyway, do it at the start. There is a chance you could skip the rest of the program.

book1978 on April 7, 2023, 11 p.m.
Надеюсь это решение вам больше понравится :) <pre class='brush: python'>def words_order(text: str, words: list) -> bool: text = text.split() s = [text.index(i) + 1 if i in text else 0 for i in words] return all(s) and len(words) == len(set(words)) and s == sorted(s) </pre>

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