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C'est la Z

Bracket Bonanza (AOC 2021 day 10)

I know, where's day 9? Thursdays (and Mondays) are already tight for me - I teach all morning and it's been a rough week. I just had very little energy and focus all day yesterday. I snuck some time in to finish part 1 but couldn't focus on part 2.

Today, however, my body gave me an extra half hour of sleep (til 4:30am) so I had extra time and energy. I was fortunate in that today's problem was essentially one that I've assigned time and time again so I was able to wrap it up quickly and then go back to finish day 9.

It's pretty likely that there'll be a day I can't solve in the near future so maybe I'll write up day 9 then.

But for now, day 10 - Syntax Scoring or as I like to say a bracket bonanza. Inputs were lines of brackets like this:


Both parts of the problem involve figuring out if the line is valid, that is, if each open bracket has a matching close bracket without overlap. For example (([])) is valid but (([)]) is not becaue the inner () and the [] overlap.

I could be like a kid doing a technical interview and pretend I hadn't solved the problem before the interview but the truth is, I've done problems very similar to this most times I teach data structures.

As you process through the input, whenever you see a closing bracket, it has to match up with the last seen open bracket otherwise the expression is invalid. This means you have to maintain a data structure in such a way that the last open bracket you see is the first one you check whenever you see a closing brace.

This is, by definition, a stack. A stack is a data structure where the last item you put in is the first item you take out.

So, the algorithm is pretty simple.

make an empty stack
For each character in the input:
  1. If it's an open bracket put it on the stack
  2. If it's a close bracket, check the top of the stack.
     1. If the stack is empty you've got an invalid expression - exit
     2. If top of the stack doesn't match your bracket - invalid -exit
     3. If top of the stack does match your braket, pop it off the stack and keep going

That's basically it. The only remaining task is to figure out the specifics to answer the problem.

For part 1 you had to run the above algorithm on each line of input and for each invalid expression look up the invalid character's value (provide in the problem text) and add them all up.

For part two you had to take the remaining characters in the lines once you hit an invalid brace and figure out a slightly more complex score.

So, given that I had basically done this problem it wasn't hard but it was fun. This could be directly assigned to any data structures class.

As usual, my code can be found here:

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