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First solution in Clear category for Acceptable Password III by jnruby
# Taken from mission Acceptable Password II
# Taken from mission Acceptable Password I
def is_acceptable_password(password: str) -> bool:
# your code here
if len(password) < 6:
return False
if not any(char.isalpha() for char in password):
return False
if not any(char.isdigit() for char in password):
return False
else: return True
Dec. 14, 2020
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