Naive, does the job solution in Clear category for House Password by Telkandore
#The password will be considered strong enough if its length is greater
#than or equal to 10 symbols, it has at least one digit, as well as
#containing one uppercase letter and one lowercase letter in it.
#The password contains only ASCII latin letters or digits.
if len(data) < 10 :
hasDigit = hasUpper = hasLower = False
for letter in data :
if letter.islower() :
hasLower = True
if letter.isupper() :
hasUpper = True
if letter.isdigit() :
hasDigit = True
return hasDigit and hasUpper and hasLower
#Just check all conditions
if __name__ == '__main__':
#These "asserts" using only for self-checking and not necessary for auto-testing
assert checkio('A1213pokl') == False, "1st example"
assert checkio('bAse730onE4') == True, "2nd example"
assert checkio('asasasasasasasaas') == False, "3rd example"
assert checkio('QWERTYqwerty') == False, "4th example"
assert checkio('123456123456') == False, "5th example"
assert checkio('QwErTy911poqqqq') == True, "6th example"
July 16, 2015