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First solution in Clear category for Acceptable Password VI by koladen
# Taken from mission Acceptable Password V
import re
def is_acceptable_password(a):
# your code here
return bool(re.search(r'((?=.{6,})(?=.*\d.*)(?=.*\D.*)|(?=.{10,}))', a)) and bool(re.search(r'^((?!password).)*$', a, re.I)) and len(set(a)) > 2
Oct. 30, 2020
