Day 1: Historian Hysteria

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://blocks.programming.dev/ if you prefer sending it through a URL

FAQ

  • rwdf@lemmy.world
    link
    fedilink
    arrow-up
    0
    ·
    12 days ago

    Go

    package main
    
    import (
    	"bufio"
    	"fmt"
    	"os"
    	"sort"
    	"strconv"
    	"strings"
    )
    
    func main() {
    	input, _ := os.Open("input.txt")
    	defer input.Close()
    
    	left, right := []int{}, []int{}
    
    	scanner := bufio.NewScanner(input)
    	for scanner.Scan() {
    		line := scanner.Text()
    		splitline := strings.Split(line, "   ")
    		l, _ := strconv.Atoi(splitline[0])
    		r, _ := strconv.Atoi(splitline[1])
    		left, right = append(left, l), append(right, r)
    	}
    
    	fmt.Printf("part 1 - total diff: %d\n", part1(left, right))
    	fmt.Printf("part 2 - new total: %d\n", part2(left, right))
    }
    
    func part1(left, right []int) int {
    	diff := 0
    	sort.Ints(left)
    	sort.Ints(right)
    
    	for i, l := range left {
    		if l > right[i] {
    			diff += (l - right[i])
    		} else {
    			diff += (right[i] - l)
    		}
    	}
    	return diff
    }
    
    func part2(left, right []int) int {
    	newTotal := 0
    
    	for _, l := range left {
    		matches := 0
    		for _, r := range right {
    			if l == r {
    				matches++
    			}
    		}
    		newTotal += l * matches
    	}
    	return newTotal
    }
    
  • lwhjp@lemmy.sdf.org
    link
    fedilink
    arrow-up
    0
    ·
    15 days ago

    Haskell

    Plenty of scope for making part 2 faster, but I think simple is best here. Forgot to sort the lists in the first part, which pushed me waaay off the leaderboard.

    import Data.List
    
    main = do
      [as, bs] <- transpose . map (map read . words) . lines <$> readFile "input01"
      print . sum $ map abs $ zipWith (-) (sort as) (sort bs)
      print . sum $ map (\a -> a * length (filter (== a) bs)) as
    
  • mykl@lemmy.world
    link
    fedilink
    arrow-up
    0
    ·
    14 days ago

    Uiua

    For entertainment purposes only, I’ll be trying a solution in Uiua each day until it all gets too much for me…

    $ 4   3
    $ 2   5
    $ 1   3
    $ 3   9
    $ 3   3
    ⊜∘⊸≠@\n     # Partition at \n.
    ⊜(⍆∵⋕)⊸≠@\s # Partition at space, parse ints, sort.
    /+/(⌵-)     # Get abs differences, sum.
    
  • Sleepless One@lemmy.ml
    link
    fedilink
    English
    arrow-up
    0
    ·
    15 days ago

    I’m doing it in Rust again this year. I stopped keeping up with it after day 3 last year, so let’s hope I last longer this time around.

    Solution Spoiler Alert
    use std::collections::HashMap;
    
    use crate::utils::read_lines;
    
    pub fn solution1() {
        let (mut id_list1, mut id_list2) = get_id_lists();
    
        id_list1.sort();
        id_list2.sort();
    
        let total_distance = id_list1
            .into_iter()
            .zip(id_list2)
            .map(|(left, right)| (left - right).abs())
            .sum::<i32>();
    
        println!("Total distance = {total_distance}");
    }
    
    pub fn solution2() {
        let (id_list1, id_list2) = get_id_lists();
    
        let id_count_map = id_list2
            .into_iter()
            .fold(HashMap::<_, i32>::new(), |mut map, id| {
                *map.entry(id).or_default() += 1i32;
    
                map
            });
    
        let similarity_score = id_list1
            .into_iter()
            .map(|id| id * id_count_map.get(&id).copied().unwrap_or_default())
            .sum::<i32>();
    
        println!("Similarity score = {similarity_score}");
    }
    
    fn get_id_lists() -> (Vec<i32>, Vec<i32>) {
        read_lines("src/day1/input.txt")
            .map(|line| {
                let mut ids = line.split_whitespace().map(|id| {
                    id.parse::<i32>()
                        .expect("Ids from input must be valid integers")
                });
    
                (
                    ids.next().expect("First Id on line must be present"),
                    ids.next().expect("Second Id on line must be present"),
                )
            })
            .unzip()
    }
    

    I’m keeping my solutions up on GitHub.

  • janAkali@lemmy.one
    link
    fedilink
    English
    arrow-up
    0
    ·
    edit-2
    15 days ago

    Nim

    I’ve got my first sub-1000 rank today (998 for part 2). Yay!
    Simple and straightforward challenge, very fitting for 1st day. Gonna enjoy it while it lasts.

    proc solve(input: string): AOCSolution[int, int] =
      var l1,l2: seq[int]
      for line in input.splitLines():
        let pair = line.splitWhitespace()
        l1.add parseInt(pair[0])
        l2.add parseInt(pair[1])
      l1.sort()
      l2.sort()
    
      block p1:
        for i in 0..l1.high:
          result.part1 += abs(l1[i] - l2[i])
    
      block p2:
        for n in l1:
          result.part2 += n * l2.count(n)
    

    Codeberg repo

  • mcmodknower@programming.dev
    link
    fedilink
    arrow-up
    0
    ·
    15 days ago

    Solution in ruby

    This is my third program in ruby after the ruby tutorial and half of the rails tutorial, so don’t expect anything too good from it.

    Also i did this today since i had time, i will probably not comment every day.

    • morrowind@lemmy.ml
      link
      fedilink
      arrow-up
      0
      ·
      14 days ago

      fyi for lines 14-22 you an use .abs instead of checking for negatives and .sum instead of doing it manually. Check my crystal solution to see what I mean

  • hades@lemm.ee
    link
    fedilink
    arrow-up
    0
    ·
    15 days ago

    C#

    public class Day01 : Solver
    {
      private ImmutableArray<int> left;
      private ImmutableArray<int> right;
    
      public void Presolve(string input)
      {
        var pairs = input.Trim().Split("\n").Select(line => Regex.Split(line, @"\s+"));
        left = pairs.Select(item => int.Parse(item[0])).ToImmutableArray();
        right = pairs.Select(item => int.Parse(item[1])).ToImmutableArray();
      }
    
      public string SolveFirst() => left.Sort().Zip(right.Sort()).Select((pair) => int.Abs(pair.First - pair.Second)).Sum().ToString();
    
      public string SolveSecond() => left.Select((number) => number * right.Where(v => v == number).Count()).Sum().ToString();
    }
    
  • Gobbel2000@programming.dev
    link
    fedilink
    arrow-up
    0
    ·
    15 days ago

    Rust

    Right IDs are directly read into a hash map counter.

    use std::str::FromStr;
    use std::collections::HashMap;
    
    fn part1(input: String) {
        let mut left = Vec::new();
        let mut right = Vec::new();
        for line in input.lines() {
            let mut parts = line.split_whitespace()
                .map(|p| u32::from_str(p).unwrap());
            left.push(parts.next().unwrap());
            right.push(parts.next().unwrap());
        }
        left.sort_unstable();
        right.sort_unstable();
        let diff: u32 = left.iter().zip(right)
            .map(|(l, r)| l.abs_diff(r))
            .sum();
        println!("{diff}");
    }
    
    fn part2(input: String) {
        let mut left = Vec::new();
        let mut right: HashMap<u32, u32> = HashMap::new();
        for line in input.lines() {
            let mut parts = line.split_whitespace()
                .map(|p| u32::from_str(p).unwrap());
            left.push(parts.next().unwrap());
            *right.entry(parts.next().unwrap()).or_default() += 1;
        }
        let similar: u32 = left.iter()
            .map(|n| n * right.get(n).copied().unwrap_or_default())
            .sum();
        println!("{similar}");
    }
    
    util::aoc_main!();
    
  • proved_unglue@programming.dev
    link
    fedilink
    arrow-up
    0
    ·
    12 days ago

    Kotlin

    No 💜 for Kotlin here?

    import kotlin.math.abs
    
    fun part1(input: String): Int {
        val diffs: MutableList<Int> = mutableListOf()
        val pair = parse(input)
        pair.first.sort()
        pair.second.sort()
        pair.first.forEachIndexed { idx, num ->
            diffs.add(abs(num - pair.second[idx]))
        }
        return diffs.sum()
    }
    
    fun part2(input: String): Int {
        val pair = parse(input)
        val frequencies = pair.second.groupingBy { it }.eachCount()
        var score = 0
        pair.first.forEach { num ->
            score += num * frequencies.getOrDefault(num, 0)
        }
        return score
    }
    
    private fun parse(input: String): Pair<MutableList<Int>, MutableList<Int>> {
        val left: MutableList<Int> = mutableListOf()
        val right: MutableList<Int> = mutableListOf()
        input.lines().forEach { line ->
            if (line.isNotBlank()) {
                val parts = line.split("\\s+".toRegex())
                left.add(parts[0].toInt())
                right.add(parts[1].toInt())
            }
        }
        return left to right
    }
    
    • the_beber@lemm.ee
      link
      fedilink
      English
      arrow-up
      0
      ·
      edit-2
      12 days ago

      I have another Kotlin (albeit similar) solution:

      import kotlin.math.abs
      
      fun main() {
      
          fun getLists(input: List<String>): Pair<List<Int>, List<Int>> {
              val unsortedPairs = input.map {
                  it.split("   ").map { it.toInt() }
              }
      
              val listA = unsortedPairs.map { it.first() }
              val listB = unsortedPairs.map { it.last() }
              return Pair(listA, listB)
          }
      
          fun part1(input: List<String>): Int {
              val (listA, listB) = getLists(input)
      
              return listA.sorted().zip(listB.sorted()).sumOf { abs(it.first - it.second) }
          }
      
          fun part2(input: List<String>): Int {
              val (listA, listB) = getLists(input)
      
              return listA.sumOf { number ->
                  number * listB.count { it == number }
              }
          }
      
          // Or read a large test input from the `src/Day01_test.txt` file:
          val testInput = readInput("Day01_test")
          check(part1(testInput) == 11)
          check(part2(testInput) == 31)
      
          // Read the input from the `src/Day01.txt` file.
          val input = readInput("Day01")
          part1(input).println()
          part2(input).println()
      }
      
      

      It’s a bit more compact. (If you take out the part that actually calls the functions on the (test-)input.)

  • aurele@sh.itjust.works
    link
    fedilink
    arrow-up
    0
    ·
    10 days ago

    Elixir

    defmodule AdventOfCode.Solution.Year2024.Day01 do
      use AdventOfCode.Solution.SharedParse
    
      @impl true
      def parse(input) do
        numbers =
          input
          |> String.split("\n", trim: true)
          |> Enum.map(fn l -> String.split(l, ~r/\s+/) |> Enum.map(&String.to_integer/1) end)
    
        {Stream.map(numbers, &Enum.at(&1, 0)), Stream.map(numbers, &Enum.at(&1, 1))}
      end
    
      def part1({left, right}) do
        Enum.zip_reduce(Enum.sort(left), Enum.sort(right), 0, &(&3 + abs(&1 - &2)))
      end
    
      def part2({left, right}) do
        freq = Enum.frequencies(right)
        left |> Stream.map(&(&1 * Map.get(freq, &1, 0))) |> Enum.sum()
      end
    end
    
  • sjmulder@lemmy.sdf.org
    link
    fedilink
    arrow-up
    0
    ·
    15 days ago

    Solution in C

    Part 1 is a sort and a quick loop. Part 2 could be efficient with a lookup table but it was practically instant with a simple non-memoized scan so left it that way.

    • mcmodknower@programming.dev
      link
      fedilink
      English
      arrow-up
      0
      ·
      edit-2
      15 days ago

      You are using some interesting techniques there. I never imaged you could use the result of == for adding to a counter.

      But how did you handle duplicates in part 2?

  • MalachiAzrael@sh.itjust.works
    link
    fedilink
    English
    arrow-up
    0
    ·
    14 days ago

    Python

    Part 1
    left_list = []
    right_list = []
    
    for line in file:
        split_line = line.split()
        left_list.append(int(split_line[0]))
        right_list.append(int(split_line[1]))
    
    sorted_left = sorted(left_list)
    sorted_right = sorted(right_list)
    distance = []
    for left, right in zip(sorted_left, sorted_right):
        distance.append(abs(left - right))
    
    total = sum(distance)
    
    print(total)
    
    Part 2
    file = open('input.txt', 'r')
    left_list = []
    right_list = []
    
    for line in file:
        split_line = line.split()
        left_list.append(int(split_line[0]))
        right_list.append(int(split_line[1]))
    
    sim_score = 0
    for item in left_list:
        sim = right_list.count(item)
        sim_score += (sim * item)
    
    print(sim_score)
    

    I am sure there were better ways to do this, this was just the first way in my head, in the order it appeared

  • Leavingoldhabits@lemmy.world
    link
    fedilink
    arrow-up
    0
    ·
    14 days ago

    I’m quite inexperienced as a programmer, I learned most of the basic concepts from playing human resource machine and 7 billion humans. After mucking about writing some CLI utilities in Perl and python, I’ve decided to give rust a go.

    Part 1

    Part 2

  • Rin@lemm.ee
    link
    fedilink
    arrow-up
    0
    ·
    edit-2
    14 days ago

    TypeScript

    Solution
    import { AdventOfCodeSolutionFunction } from "./solutions";
    
    function InstancesOf(sorted_array: Array<number>, value: number) {
        const index = sorted_array.indexOf(value);
        if(index == -1)
            return 0;
    
        let sum = 1;
    
        for (let array_index = index + 1; array_index < sorted_array.length; array_index++) {
            if(sorted_array[array_index] != value)
                break;
    
            sum += 1;
        }
    
        return sum;
    }
    
    export const solution_1: AdventOfCodeSolutionFunction = (input) => {
        const left: Array<number> = [];
        const right: Array<number> = [];
    
        const lines = input.split("\n");
    
        for (let index = 0; index < lines.length; index++) {
            const element = lines[index].trim();
            if(!element)
                continue;
    
            const leftRight = element.split("   ");
            left.push(Number(leftRight[0]));
            right.push(Number(leftRight[1]));
        }
    
        const numSort = (a: number, b: number) => a - b;
        left.sort(numSort);
        right.sort(numSort);
    
        let sum = 0;
        for (let index = 0; index < left.length; index++) {
            const leftValue = left[index];
            const rightValue = right[index];
    
            sum += Math.abs(leftValue - rightValue);
        }
    
        const part1 = `Part 1: ${sum}`;
    
        sum = 0;
        for (let index = 0; index < left.length; index++) {
            sum += left[index] * InstancesOf(right, left[index]);
        }
    
        const part2 = `Part 2: ${sum}`;
    
        return `${part1}\n${part2}`;
    };
    

    Not the most elegant solution but it works. Decided to reuse the array since it is sorted for both sides.

  • Euro@programming.dev
    link
    fedilink
    arrow-up
    0
    ·
    14 days ago

    Viml

    I think viml is a very fun language, i like weird languages lol, so this year im doing it in viml while trying to use as many of the original ed/ex commands as i can (:d, :p, :a, :g, …)

    Part 1
    !cp ./puzzle1 ./puzzle1.editing
    e ./puzzle1.editing
    
    1,$sort
    let row1 = []
    
    g/^\d/let row1 = add(row1, str2nr(expand("<cword>"))) | norm 0dw
    1d
    1,$sort
    g/^\d/execute 'norm cc' .. string(abs(expand("<cword>") - row1[line('.') - 1]))
    
    $a|---ANSWER---
    0
    .
    1,$-1g/^\d/call setline("$", str2nr(getline("$")) + str2nr(expand("<cword>")))
    
    Part 2
    read ./puzzle1
    
    let cnt = 0
    g/^\d/let cnt += expand("<cword>") *
                \ searchcount(#{pattern: '\s\+' .. expand("<cword>")}).total
    
    echo cnt .. "\n"
    
    w! ./puzzle1.editing