this post was submitted on 03 Dec 2024
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Advent Of Code

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An unofficial home for the advent of code community on programming.dev!

Advent of Code is an annual Advent calendar of small programming puzzles for a variety of skill sets and skill levels that can be solved in any programming language you like.

AoC 2024

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Day 3: Mull It Over

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[โ€“] [email protected] 5 points 1 day ago (2 children)

I couldn't figure it out in haskell, so I went with bash for the first part

Shell

cat example | grep -Eo "mul\([[:digit:]]{1,3},[[:digit:]]{1,3}\)" | cut -d "(" -f 2 | tr -d ")" | tr "," "*" | paste -sd+ | bc

but this wouldn't rock anymore in the second part, so I had to resort to python for it

Python

import sys

f = "\n".join(sys.stdin.readlines())

f = f.replace("don't()", "\ndon't()\n")
f = f.replace("do()", "\ndo()\n")

import re

enabled = True
muls = []
for line in f.split("\n"):
    if line == "don't()":
        enabled = False
    if line == "do()":
        enabled = True
    if enabled:
        for match in re.finditer(r"mul\((\d{1,3}),(\d{1,3})\)", line):
            muls.append(int(match.group(1)) * int(match.group(2)))
        pass
    pass

print(sum(muls))
[โ€“] [email protected] 2 points 1 day ago

Really cool trick. I did a bunch of regex matching that I'm sure I won't remember how it works few weeks from now, this is so much readable

[โ€“] proved_unglue 1 points 1 day ago

Nice, sometimes a few extra linebreaks can do the trick...