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len if isdigit solution in Clear category for Acceptable Password VI by m.kurapov
def is_acceptable_password(password: str) -> bool:
return password.lower().find('password')==-1 \
and (len(password)>9
or
len(password)>6 and \
len(password)>len([x for x in password if x.isdigit()])>0
) \
and len(set(password))>2
Nov. 18, 2020
