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
- What is this?: Here is a post with a large amount of details: https://programming.dev/post/22323136
- Where do I participate?: https://adventofcode.com/
- Is there a leaderboard for the community?: We have a programming.dev leaderboard with the info on how to join in this post: https://programming.dev/post/6631465
Uiua
For entertainment purposes only, I’ll be trying a solution in Uiua each day until it all gets too much for me…
$ 3 4 $ 4 3 $ 2 5 $ 1 3 $ 3 9 $ 3 3 ⊜∘⊸≠@\n # Partition at \n. ⊜(⍆∵⋕)⊸≠@\s # Partition at space, parse ints, sort. &p/+/(⌵-). # Part1 : Get abs differences, sum, print. &p/+×⟜(/+⍉≡⌕)°⊂ # Part 2 : Count instances, mul out, sum, print.
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
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)
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!();
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.
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 }
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.)
Thanks! I like the
Pair
destruction andzip().sumOf()
approach. I’m relatively new to Kotlin, so this is a good learning experience. 😅
Smalltalk
day1p12: input | list1 list2 nums dist sim | list1 := OrderedCollection new. list2 := OrderedCollection new. input linesDo: [ :l | nums := l substrings collect: [ :n | n asInteger ]. list1 add: (nums at: 1). list2 add: (nums at: 2). ]. list1 sort. list2 sort. dist := 0. list1 with: list2 do: [ :a :b | dist := dist + (a - b) abs ]. sim := list1 sumNumbers: [ :x | x * (list2 occurrencesOf: x) ]. ^ Array with: dist with: sim.
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.
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?
I’m not sure if I understand the question correctly but for every number in the left array I count in the right array. That means duplicate work but shrug 😅
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(); }
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
Not going to push hard on these first days (fever being a reason), so I slept in quite a bit before looking at the problem.
C#
List<int> _LeftList = new List<int>(); List<int> _RightList = new List<int>(); // Fed via File.ReadLines(...).Select(l => l.Trim()) public void Input(IEnumerable<string> lines) { foreach (var line in lines) { var split = line.Split(' ', StringSplitOptions.RemoveEmptyEntries).Select(s => int.Parse(s)); _LeftList.Add(split.First()); _RightList.Add(split.Last()); } } public void Part1() { Console.WriteLine($"Sum: {_LeftList.Order().Zip(_RightList.Order()).Select(v => Math.Abs(v.First - v.Second)).Sum()}"); } public void Part2() { Console.WriteLine($"Sum: {_LeftList.Select(l => _RightList.Where(i => i == l).Count() * l).Sum()}"); }
I’m late to the party, as usual. Damned timezones. This year I’m going to tackle with a small handful of languages, but primarily Elixir and Gleam. This is my first time trying this languages in earnest, so expect some terrible, inefficient and totally unidiomatic code!
Here’s day one:Elixir
part_one = File.read!("input.in") |> String.split("\n", trim: true) |> Enum.map(fn line -> line |> String.split() |> Enum.map(&String.to_integer/1) end) |> Enum.reduce({[], []}, fn [first, second], {list1, list2} -> {[first | list1], [second | list2]} end) |> then(fn {list1, list2} -> {Enum.sort(list1), Enum.sort(list2)} end) |> then(fn {list1, list2} -> Enum.zip(list1, list2) |> Enum.map(fn {x, y} -> abs(x - y) end) end) |> Enum.sum() part_two = File.read!("input.in") |> String.split("\n", trim: true) |> Enum.map(fn line -> line |> String.split() |> Enum.map(&String.to_integer/1) end) |> Enum.reduce({[], []}, fn [first, second], {list1, list2} -> {[first | list1], [second | list2]} end) |> then(fn {list1, list2} -> Enum.map(list1, fn line -> line * Enum.count(list2, fn x -> x === line end) end) |> Enum.sum() end) IO.inspect(part_one) IO.inspect(part_two)
Rust
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() }
Raku
I’m trying warm up to Raku again.
Solution
use v6; sub MAIN($input) { my $file = open $input; grammar LocationList { token TOP { <row>+%"\n" "\n"* } token row { <left=.id> " "+ <right=.id> } token id { \d+ } } my $locations = LocationList.parse($file.slurp); my @rows = $locations<row>.map({ (.<left>.Int, .<right>.Int)}); my $part-one-solution = (@rows[*;0].sort Z- @rows[*;1].sort)».abs.sum; say "part 1: $part-one-solution"; my $rbag = bag(@rows[*;1].sort); my $part-two-solution = @rows[*;0].map({ $_ * $rbag{$_}}).sum; say "part 2: $part-two-solution"; }
I’m happy to see that Lemmy no longer eats Raku code.