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Acceptable Password 3 by Ola_N solution in Uncategorized category for Acceptable Password III by Aleksandra_Niewiadomska
def is_acceptable_password(password: str) -> bool:
return len(password) > 6 and any(str.isdigit() for str in password) and not password.isdigit()
assert is_acceptable_password("short") == False
assert is_acceptable_password("muchlonger") == False
assert is_acceptable_password("ashort") == False
assert is_acceptable_password("muchlonger5") == True
assert is_acceptable_password("sh5") == False
assert is_acceptable_password("1234567") == False
print("The mission is done! Click 'Check Solution' to earn rewards!")
Nov. 28, 2022
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