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Regex solution in Clear category for Acceptable Password V by H0r4c3
import re
def is_acceptable_password(password: str) -> bool:
if len(password) > 6 and (re.findall('[\D]+[\d]+', password) or len(password) > 9) and 'password' not in password.lower():
return True
else:
return False
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
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Nov. 16, 2021
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