First clear solution for Acceptable Password V by dx2-66 - python coding challenges

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First solution in Clear category for Acceptable Password V by dx2-66

import re
def is_acceptable_password(passw):
    match = re.search(r'\d+', passw)
    match2 = re.search(r'password', passw, re.IGNORECASE)
    n_digits = len(match[0]) if match else 0
    return not match2 and (len(passw) > 9 or n_digits and n_digits < len(passw) > 6)

May 6, 2022

Comments:

CDG.Axel on March 16, 2023, 1:31 p.m.
<p>Imo re is cool sometimes, but not for that. And "0 < n<em>digits <" shorter than "n</em>digits and n_digits <"</p> <pre class='brush: python'>match = re.search(r'\d+', passw) match2 = re.search(r'password', passw, re.IGNORECASE) n_digits = len(match[0]) if match else 0 return not match2 and (len(passw) > 9 or n_digits and n_digits < len(passw) > 6) --> n_digits = sum(map(str.isdigit, passw)) match2 = passw.lower().find('password') < 0 # < 0 for avoid 'not match2' in condition return match2 and (len(passw) > 9 or 0 < n_digits < len(passw) > 6) </pre>

Strings Theory

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