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First solution in Clear category for Acceptable Password V by tokiojapan55
def is_acceptable_password(password: str) -> bool:
if len(password) <= 6:
return False
if password.lower().find('password') >= 0:
return False
if len(password) >= 10:
return True
nums = [c for c in password if c.isdigit()]
if len(nums) == 0 or len(nums) == len(password):
return False
return True
June 23, 2020
