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Checking for CharactersNow that you're eliminating the most common passwords, you can add some more sophisticated checks to your algorithm. A good policy is to disallow plain dictionary words. If you require the user to include nonalphabetic characters, there'll be some amount of randomness even if the password includes dictionary words (such as "apple1984"). You might also require passwords be a minimum length. The longer a password, the longer it takes for a hacker to try all possibilities. For example, there are only 140,608 simple six-letter passwords. In contrast, the number of 16-character passwords generated according to the rules below has an upper bound of over 30 nonillion, or 30 thousand billion billion billion! That's way too many for a hacker to perform an exhaustive search of possible passwords.Use the following rules:At least 16 charactersAt least one regular letterAt least one digitAt least one punctuation characterlet tenMostCommonPasswords = [ "123456", "password", "12345678", "qwerty", "12345", "123456789", "letmein", "1234567", "football", "iloveyou"]let digits = "0123456789"let punctuation = "!@#$%^&*(),.<>;'`~[]{}\\|/?_-+= "Implement your updated algorithm below.To implement your algorithm, you can treat the password as an array of characters. Use a for...in loop to examine all of its characters and to keep a count of digits and punctuation characters. (You can use the contains() method you used in ArraysAndLoops.playground to help with this.) After checking all the characters in the password, write a final conditional statement to check whether you found at least one of each type of required character.How do I create the correct algorithm using a for..in loop and the contains() method?this is what I have tried:let password = "password"password.countfunc correctAuthenticationMethod(ofPassword: String) -> Bool { for char in password { if digits.contains(char) { if punctuation.contains(char) { return true } else { return false } } else { return true } } return false}correctAuthenticationMethod(ofPassword: password)I appreciate any help with this issue 🙂