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Simple solution in Clear category for Acceptable Password III by mindaugas.dadurkevicius
def is_acceptable_password(password):
len_test = len(password) > 6
digit_test1 = max([ch.isdigit() for ch in password])
digit_test2 = not password.isdigit()
return len_test & digit_test1 & digit_test2
June 18, 2021
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