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simple solution in Clear category for Acceptable Password V by Bernardo_Pereira
def is_acceptable_password(password: str) -> bool:
numlen = 0
if 'password' in password.lower():
return False
if len(password) <= 6:
return False
for i in range(len(password)):
if password[i].isdigit():
numlen += 1
return len(password) > 9 or (numlen > 0 and numlen < len(password))
Oct. 1, 2021