Simple clear solution for Goes Right After by thealfest1 - python coding challenges

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Simple solution in Clear category for Goes Right After by thealfest1

def goes_after(word: str, first: str, second: str) -> bool:
    return word.find(second) - word.find(first) == 1

March 12, 2020

Comments:

Sioul on April 22, 2020, 3:08 p.m.
"If one of the symbols are not in the given word - your function should return False." Does not work when first is not in word, and second is the first letter of word: <pre class='brush: python'>>>> goes_after("word", "a", "w") True </pre>

thealfest1 on April 23, 2020, 5:27 a.m.
Thanks, they should improve tests.

leggewie on June 13, 2021, 9:38 a.m.
You are right, they should improve the tests, and they <a href="https://github.com/CheckiO-Missions/checkio-mission-goes-after/commit/a6b1cd3a28ef585bd057a44f7a52c4f0b2bf1318#diff-e782327fdc88ec62bb742e5670dcffe0ae16dcb4d317188eec8cc922ea9eec5d">did</a>. At the same time, it's also the programmer's job to carefully code.

thealfest1 on May 4, 2020, 6:06 a.m.
Corrected version: def goes_after(word: str, first: str, second: str) -> bool: return -1 < word.find(second) - 1 == word.find(first)

Strings and Integers

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