First clear solution for Acceptable Password IV by juestr - python coding challenges

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First solution in Clear category for Acceptable Password IV by juestr

def is_acceptable_password(password: str) -> bool:
    check = lambda c, f: c(map(f, password))
    return len(password) > 9 or \
        len(password) > 6 and check(any, str.isdigit) and not check(all, str.isdigit)

Aug. 6, 2021

Comments:

CDG.Axel on Jan. 20, 2022, 4:54 p.m.
Great! Btw you can count digits and use it for expression, it will be one-liner. <pre class='brush: python'>return len(password) > 9 or 0 < digit_count_expression < len(password) > 6 </pre> PS: you can count digits this way, for example <pre class='brush: python'>sum(map(str.isdigit, password)) </pre>

juestr on Jan. 20, 2022, 5:46 p.m.
Thanks! I realized there are shorter ways to do this, but decided to reuse the `check`-helper from earlier missions to show off functional composition.

Strings Theory

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