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First solution in Clear category for Acceptable Password VI by yoichi
import string
def is_acceptable_password(password: str) -> bool:
if 'password' in password.lower():
return False
if len(password) <= 6:
return False
if len(set(password)) < 3:
return False
if len(password) >= 10:
return True
if all(c in string.digits for c in password):
return False
if all(c not in string.digits for c in password):
return False
return True
July 13, 2020
