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one or no and solution in Clear category for Acceptable Password IV by CDG.Axel
def is_acceptable_password(password: str) -> bool:
return 0 < sum(map(str.isdigit, password)) < len(password) > 6 or len(password) > 9
"""
return 0 < sum([i.isdigit() for i in password]) < len(password) > 6 or len(password) > 9
return (len(password)>6 and not password.isdigit() and not password.isalpha()) or len(password)>9
"""
if __name__ == "__main__":
# These "asserts" are used for self-checking and not for an auto-testing
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
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Oct. 12, 2021
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