Advent of Code – Day 11
Contents
Welcome to day 11 of advent of code where will be helping Santa to create secure passwords. We have already been given how to create new passwords but there is a new security guy who has come up with three new requirements to make the passwords more secure. So let’s start by tackling these three requirements.
We are essentially trying to check the password for three conditions and it can easily be done by using pattern matching by regular expressions.
The 3 conditions of Security Department
Condition 1
The first condition is to find if the given password contains at least
two double pairs of letters. We will use Regex.Matches
which gives a
MatchCollection
containing all the matches present for the given
regular expression. If we have more than 2 distinct pairs, then we are
good.
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Condition 2
The second condition is to find at least 1 run of 3 or more characters. We can iterate through the password and for each character check the next two characters. we would want that the difference between the successive characters is always 1. If that’s not the case till the very last character and the condition is not satisfied, then we simply reject the password.
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Condition 3
The third condition is pretty easy to understand. We simply have to
check whether the string contains the letters i
, o
or l
and if it
is then we reject the password.
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The three conditions can be clubbed together to check the validity of the password.
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Incrementing the Password
With these three conditions being taken care of we can concentrate on creating the iterator of the password with takes in a password and returns a new password by incrementing it.
The first step in tackling this requirment is to reverse the password
because that will make on the algorithms a bit more symmetric. Then you
can start from the beginning (which would have been the end of the
original password) increment the characters from left to right. If the
first character is z
then we increment it and increment the rest of
the string. If the first character is anything other than z
then we do
not touch the rest of the string.
Since C# doesn’t provide a method to reverse a string (unless you treat
it as an IEnumerable<char>
), the following method works well for
non-unicode strings which we have for this problem. More solutions are
here.
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Next Password
The last step is to start with a password, call the Increment
function
and keep calling it unless we find valid password which passes the
IsValidPassword
check above.
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I have kept all the functions here static because none of these depend on the shared state. In fact there is no shared state here. Everything is kind of functional – it depends on whatever the argument is passed in and simply returns a new output.
The code is present at github.
Author Tushar Tyagi
LastMod Jan 27, 2017