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First solution in Clear category for Acceptable Password VI by imloafer
# Taken from mission Acceptable Password V
# Taken from mission Acceptable Password IV
# Taken from mission Acceptable Password III
# Taken from mission Acceptable Password II
# Taken from mission Acceptable Password I
import re
from collections import Counter
def is_acceptable_password(password: str) -> bool:
# your code here
return all([any([all([len(password) > 5, re.search(r'\d', password), re.search(r'\D', password)]),
len(password) > 9]), len(Counter(password).keys()) >= 3,
'password' not in password.lower()])
assert is_acceptable_password("short") == False
assert is_acceptable_password("short54") == True
assert is_acceptable_password("muchlonger") == True
assert is_acceptable_password("ashort") == False
assert is_acceptable_password("muchlonger5") == True
assert is_acceptable_password("sh5") == False
assert is_acceptable_password("1234567") == False
assert is_acceptable_password("12345678910") == True
assert is_acceptable_password("password12345") == False
assert is_acceptable_password("PASSWORD12345") == False
assert is_acceptable_password("pass1234word") == True
assert is_acceptable_password("aaaaaa1") == False
assert is_acceptable_password("aaaaaabbbbb") == False
assert is_acceptable_password("aaaaaabb1") == True
assert is_acceptable_password("abc1") == False
assert is_acceptable_password("abbcc12") == True
assert is_acceptable_password("aaaaaaabbaaaaaaaab") == False
Sept. 29, 2022
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