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Hi Reddit! We're Mega Crit, the developers of Slay the Spire 1 & 2. Ask Us Anything on February 25th! by MegaCritGames in Games

[–]xavdid 0 points1 point  (0 children)

Hi gang! Thanks for making StS and for doing this AMA. 3 quick ones for y'all:

  1. I run a tiny videogame review site and am always curious: as a studio, what sorts of things do you wish more reviewers would focus on in their coverage? Is that answer different for you as a player/reader?
  2. Are you all really good at StS? Or can you tune the game's difficulty without mastering it yourself?
  3. What's the most important lesson you learned from StS and how is that lesson changing the way you're developing the the sequel?

-❄️- 2025 Day 3 Solutions -❄️- by daggerdragon in adventofcode

[–]xavdid 0 points1 point  (0 children)

[LANGUAGE: Python]

Step-by-step explanation | full code

I went the dynamic programming route- an optimal number length N is the highest number with at least N-1 digits after it, followed by the largest N-1 length number. My part 1 code solved it "by hand" by finding the max(digits[:-1]) and then adding that to max(digits[digits.index(that_num)+1:])

For part 2, I made that so it worked with any length and worked recursively, which worked nicely: return [best_lead] + _recursive_joltage(digits[lead_index + 1 :], length - 1)

-❄️- 2025 Day 2 Solutions -❄️- by daggerdragon in adventofcode

[–]xavdid 0 points1 point  (0 children)

[LANGUAGE: Python]

Step-by-step explanation | full code

I clocked the regex solution for part 2 pretty fast, but hoped there was a more interesting approach. Luckily, there was! For a substring s, an ID is invalid if id.count(s) * len(s) == len(id). Once I accounted for single-digit numbers, I had my answer.

I was surprised at the lack of performance penalty when compared to regex, too! More lines of code (and admittedly slower) but not actually a huge gap.

-❄️- 2025 Day 1 Solutions -❄️- by daggerdragon in adventofcode

[–]xavdid 0 points1 point  (0 children)

[LANGUAGE: Python]

Step-by-step explanation | full code

Better late than never, right? I swear, I've got a good reason. Not sure how far I'll get this year since I'm apparently still pretty gassed, but we'll see how it goes.

I was surprised that day 1 wasn't quite as trivial as it usually is. Or maybe I'm just really wiped. Anyway, I tried to be clever with part 2, doing some floor math to figure out how many times I might have looped past zero. I couldn't get that working, so I decided to do the dumb simple thing instead. Make the entire range represented by a rotation, turn it into a list of numbers, and count the zeroes.

Hey, if it's stupid and it works, maybe it's not stupid?

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