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First solution in Clear category for Acceptable Password IV by Ekaterina_Hostinetska
def is_acceptable_password(password: str) -> bool:
digits = 0
for p in password:
if p.isdigit():
digits += 1
a = len(password) > digits > 0
return len(password) >= 6 and (a if len(password) <= 9 else True)
Aug. 16, 2021
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