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Acceptable Password VI solution in Uncategorized category for Acceptable Password VI by Aleksandra_Niewiadomska
def is_acceptable_password(password: str) -> bool:
if len(password) <= 6:
return False
if "password" in password.lower():
return False
if len(password) <= 9:
if password.isdigit():
return False
if not any(str.isdigit() for str in password):
return False
unique_chars = set(password)
if len(unique_chars) < 3:
return False
return True
print("Example:")
print(is_acceptable_password("short"))
# These "asserts" are used for self-checking
assert is_acceptable_password("short") == False
assert is_acceptable_password("short54") == True
assert is_acceptable_password("muchlonger") == True
assert is_acceptable_password("ashort") == False
assert is_acceptable_password("muchlonger5") == True
assert is_acceptable_password("sh5") == False
assert is_acceptable_password("1234567") == False
assert is_acceptable_password("12345678910") == True
assert is_acceptable_password("password12345") == False
assert is_acceptable_password("PASSWORD12345") == False
assert is_acceptable_password("pass1234word") == True
assert is_acceptable_password("aaaaaa1") == False
assert is_acceptable_password("aaaaaabbbbb") == False
assert is_acceptable_password("aaaaaabb1") == True
assert is_acceptable_password("abc1") == False
assert is_acceptable_password("abbcc12") == True
assert is_acceptable_password("aaaaaaabbaaaaaaaab") == False
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Feb. 1, 2023
