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First solution in Clear category for Acceptable Password V by Red_Ale
def is_acceptable_password(word):
if 'password' in word.lower(): return False
return len(word) > 9 or (len(word) > 6 and not word.isdigit() and True in [x.isdigit() for x in word])
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
assert is_acceptable_password("password12345") == False
assert is_acceptable_password("PASSWORD12345") == False
assert is_acceptable_password("pass1234word") == True
print("Coding complete? Click 'Check' to earn cool rewards!")
Nov. 30, 2021
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