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

Megathread guidelines

  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
  • You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL

FAQ

(page 2) 5 comments
sorted by: hot top controversial new old
[–] [email protected] 1 points 3 weeks ago

TypeScript

Solution

import { AdventOfCodeSolutionFunction } from "./solutions";

export const solution_3: AdventOfCodeSolutionFunction = (input) => {
    const mul_regex = /mul\((\d+),(\d+)\)/g; // mul()
    const do_regex = /do\(\)/g;              // do()
    const do_not_regex = /don\'t\(\)/g;      // don't()

    const doLength = "do()".length;
    const doNotLength = "don't()".length;

    let input_copy = "" + input;
    let part_1 = 0;
    let part_2 = 0;
    let isEnabled = true;
    while (true) {
        const nextMul = input_copy.search(mul_regex);
        const nextDo = input_copy.search(do_regex);
        const nextDoNot = input_copy.search(do_not_regex);
        let pointer = Number.POSITIVE_INFINITY;

        // find the smallest while ignoring items that are not found
        if (nextMul != -1)
            pointer = Math.min(pointer, nextMul);

        if (nextDo != -1)
            pointer = Math.min(pointer, nextDo);

        if (nextDoNot != -1)
            pointer = Math.min(pointer, nextDoNot);

        // no matches
        if (pointer == Number.POSITIVE_INFINITY)
            break

        // handle found command
        switch (pointer) {
            case nextDo: {
                pointer += doLength;
                isEnabled = true;
                break;
            }

            case nextDoNot: {
                pointer += doNotLength;
                isEnabled = false;
                break;
            }

            case nextMul: {
                const res = input_copy.matchAll(mul_regex).next().value;
                if (!res) {
                    // this should never happen but here's an escape hatch
                    throw Error("regex result is undefined or null");
                }

                // add the length of the whole capture to the pointer
                pointer += res[0].length;
                
                // multiply capture groups
                const comp = Number(res[1]) * Number(res[2]);

                // part 1 sum
                part_1 += comp;

                // part 2 sum
                if(isEnabled)
                    part_2 += comp;

                break;
            }
        }

        // shorten the start of the string
        input_copy = input_copy.slice(pointer);
    }

    return {
        part_1,
        part_2,
    };
}

This one was harder but still. I feel like I can improve it for sure :)

[–] [email protected] 1 points 2 weeks ago

I did part 2 live with the python interactive shell. I deleted all the stuff where I was just exploring ideas.

part 1:

import re

def multiply_and_add(data: "str") -> int:
    digit_matches = re.findall(r"mul\(\d{0,3},\d{0,3}\)", data)
    result = 0
    for _ in digit_matches:
        first = _.split("(")[1].split(")")[0].split(",")[0]
        second = _.split("(")[1].split(")")[0].split(",")[1]
        result += int(first) * int(second)

    return result

with open("input") as file:
    data = file.read()


answer = multiply_and_add(data)
print(answer)

part 2:

Python 3.11.2 (main, Aug 26 2024, 07:20:54) [GCC 12.2.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import solution2
<re.Match object; span=(647, 651), match='do()'>
>>> from solution2 import *
>>> split_on_dont = data.split("don't()")
>>> valid = []
>>> valid.append(split_on_dont[0])
>>> for substring in split_on_dont[1:]:
...     subsubstrings = substring.split("do()", maxsplit=1)
...     for subsubstring in subsubstrings[1:]:
...             valid.append(subsubstring)
...
>>> answer = 0
>>> for _ in valid:
...     answer += multiply_and_add(_)
...
>>> answer
103811193
[–] [email protected] 1 points 2 weeks ago

Uiua

Regex my beloved <3

Run with example input here

FindMul ← regex "mul\\((\\d+),(\\d+)\\)"

PartOne ← (
  &rs ∞ &fo "input-3.txt"
  FindMul
  /+≑(Γ—Β°βŠŸβ‹•βŠ1_2)
)

IdDont ← βŠ—β–‘"don't()"β™­

PartTwo ← (
  &rs ∞ &fo "input-3.txt"
  regex "mul\\(\\d+,\\d+\\)|do\\(\\)|don't\\(\\)"
  ⍒(IdDont.
    β†˜1βŠƒβ†˜β†™
    βŠ—β–‘"do()"β™­.
    βŠ‚β†˜1β†˜
  | IdDont.
    ≠⧻,
  )
  β–½β™­=0βŒ•β–‘"do()".
  ≑(Γ—Β°βŠŸβ‹•βŠ1_2β™­FindMul)β™­
  /+
)

&p "Day 3:"
&pf "Part 1: "
&p PartOne
&pf "Part 2: "
&p PartTwo
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