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Concise algorithm solution in Speedy category for Acceptable Password III by Igor_Sekretarev
def is_acceptable_password(password: str) -> bool:
return ( len(password) > 6
and any(ch.isdigit() for ch in password)
and not password.isdigit())
if __name__ == '__main__':
print("Example:")
print(is_acceptable_password('short'))
# These "asserts" are used for self-checking and not for an auto-testing
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("Coding complete? Click 'Check' to earn cool rewards!")
April 25, 2021
