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(Updated code, previous was removed) solution in Clear category for Acceptable Password V by thvavilis
# Taken from mission Acceptable Password IV
# Taken from mission Acceptable Password III
# Taken from mission Acceptable Password II
# Taken from mission Acceptable Password I
def is_acceptable_password(password: str) -> bool:
# your code here
if "password" in password or "PASSWORD" in password:
return False
if len(password)>9:
return True
if len(password) > 6 and not password.isdigit():
for char in password:
if char.isdigit():
return True
else:
return False
else:
return False
April 3, 2020
