Advent Of Code 2020 – DAY 2 – Object-Oriented Way6 min read

The Password Philosophy is the topic of today’s discussion. Since there is not much to do with today’s problem apart from the string parsing, I thought of doing it in an object-oriented way.

The problem statement is mostly based on the string parsing. You will get a list of lines with the password and password check policy. The task is to parse the policy and password from the string and validate that the password obeys the policy.

For example – following is the list of password policies and password,

1-3 a: abcde
1-3 b: cdefg
2-9 c: ccccccccc

Each line gives the password policy and then the password. The password policy indicates the lowest and highest number of times a given letter must appear for the password to be valid. For example, 1-3 a means that the password must contain a at least 1 time and at most 3 times.

In the above example, 2 passwords are valid. The middle password, cdefg, is not; it contains no instances of b, but needs at least 1. The first and third passwords are valid: they contain one a or nine c, both within the limits of their respective policies.

How many passwords are valid according to their policies?

So let’s see how can we solve this problem.

PART 1 – Classes & Algorithm

The task at hand are:

• Parse the line to separate out password policy and the password.
• Check if the given password obeys the policy

Here we have two clearly defined entities.

• Password
• Policy

The relation is also pretty clear. We apply the policy to password to check whether the password is correct or not. Password validity could be an attribute of a password but policy is the one to decide whether a given password is valid or not.

Another thing is that if policy changes there should be no need for a password to change. Another word, same password could become valid if a different policy operates on it.

So, with this understanding, let’s do some coding.

Password Class

Let’s start by creating a Password class. This class is very simple, it will hold the password and attribute that will tell whether the password is valid or not.

There’s one more method to find the count of the given character in the password.

Policy Class

This class is responsible for the rules that comes with the policy. It knows how to parse the information from the passed string. It also know how to validate whether a given password is valid or not.

And if password doesn’t adheres to this policy then it would be marked invalid.

Here’s the Policy class.

Another thing that we need is a file parser. But that we will do in the file itself (for simplicity).

And as we parse the line, we will be creating policy and password and applying the policy to the password to check if the password is valid.

If the password is valid then we will increment the count by one or else not.

So, this is how we can do that:

After running this code on the file we get the correct output: 572

Let’s see if this count is correct or not.

Viola!!!

The count is correct and elves are happy 🙂

Let’s move to the second part of this question.

PART – 2

Seems like the Shopkeeper made a mistake in telling us the password policy. He told us something from his prior workplace but now he checked and came up with the correct policy.

And the new policy goes like this:

Each policy actually describes two positions in the password, where 1 means the first character, 2 means the second character, and so on. (Be careful; Toboggan Corporate Policies have no concept of “index zero”!) Exactly one of these positions must contain the given letter. Other occurrences of the letter are irrelevant for the purposes of policy enforcement.

Given the same example list from above:

• 1-3 a: abcde is valid: position 1 contains a and position 3 does not.
• 1-3 b: cdefg is invalid: neither position 1 nor position 3 contains b.
• 2-9 c: ccccccccc is invalid: both position 2 and position 9 contain c.

How many passwords are valid according to the new interpretation of the policies?

Time for creating a new password policy

Since we have everything in place already, the only difference is in the policy so that is the only thing that we will change in our program.

Let’s create a new Password Policy then. Let’s name it NewPolicy and extend it from Policy.

Let’s try running the code with this new policy in place.

For that we just have to make change in one place, i.e,

policy = Policy(parts[0])

policy = NewPolicy(parts[0])

The rest of the code will stay the same.

After running the program I got the correct answer. Finally, all the elves are happy now.

Here’s the entire code:

Let me know if you have any questions or you want to discuss anything in the comments below. And don’t forget to like and share.