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First solution in Clear category for Acceptable Password VI by yan.tsarevsky
def is_acceptable_password(password):
a = len(password) > 9
b = len(password) >= 6
v = len(set(password)) >= 3
x = "password" not in password.lower()
y = bool([el for el in password if el.isdigit()])
z = bool([el for el in password if el.isalpha()])
return ((b and y and z) or a) and x and v
April 8, 2021
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