It's that time again. Advent of Code 2023. A new two part programming problem each day from now through December 25th.

Since 2015 I've looked forward to the event, taken part, and posted at least a few write ups here on my blog. I figured this year would be the same.

Of course, it does end up putting a crunch in my routine. Usually each morning I wake up, work out, usually a run unless the weather's bad, have my coffee, do the puzzle, read blogs and other content over breakfast and then I'm ready for the rest of my day.

Advent of Code adds another morning activity. Actually, maybe not this year since I broke a toe last week and am not allowed to work out for the foreseeable future.

So, this morning, I woke up, made coffee, worked on the puzzle (not quite done yet) and then looked at today's problem.

To be honest, I was a little surprised. After reading it it seemed somewhat harder than past day 1 questions. It might have to do with the fact that I haven't programmed at all since last years event and really not much in the past couple of years but still.

Usually the first days problem involves some straight through processing of the data without a whole lot of thought. After I completed today's problem I looked back on past years adn actually, while this one is a bit more involved, it really isn't that different.

Part 1 of today's problem was a straightforward as past year. You've got a bunch of lines each with some digits interspersed:

1bcc2
a1b2c3fgh4
etc.

You have to find the first and last digits in each line and take them as a number. For the above you'd have 12 for the first line and 14 for the second. You then had to sum all of those numbers up.

I thought the most straightforward way to do this was to go through the data and keep only the digits.

While I wrote my solution in Clojure, in Python for each line you could use an expression like this:

[x for x in l if x>'0' and x<='9']

Then you can just take the first and last items in the list, put them together and convert to a number:

int(x[0]+x[-1])

Finally, add all the lines together and you're done.

Part 2 threw a wrench in the works.

Now, in addition to the digits counting as digits, spelled out digits (one, two, etc.) had to count as digits as well. Looking at the sample data, no problem:

two1nine
eightwothree
abcone2threexyz
xtwone3four
4nineeightseven2
zoneight234
7pqrstsixteen

One could just replace all occurences of "one" with "1," "two" with "2," and so on.

The catch was that you could also have overlapping numbers:

9oneighthree9

Should that be converted to "91839," "91igh39," "9on8hree9," or "91839."

This made things trickier - first, how do you interpret it and then how do you solve.

It turned out that you had to allow for overlapping words so the above example would translate to 91839.

My solution, in Clojure, which you can find here involves using overlapping regular expressions - I didn't know about them so it took a while to figure that out.

Another approach, at least I think would be to replace the strings one at a time (first do the "one" occurences, then "two" etc.) but replace them with the digit within the original text. So, "one" would be replaced by "o1ne" or something similar. This way, if the final "e" was next to an "ight" you'd get that "eight" when you replaced for "eight."

All of this isn't too bad but probably required more thought than your typical advent of code day 1.

Still, a fun morning activity.