NotAwfulTech

#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
#include <set>
#include <iterator>

int main() {
  std::multiset<int> l, r;
  int a, b;
  while (std::cin >> a >> b) {
    l.insert(a); r.insert(b);
  }
  std::vector<int> delta;
  std::transform(l.begin(), l.end(), r.begin(), std::back_inserter(delta),
    [](int x, int y) { return std::abs(x-y); }
  );
  std::cout << std::accumulate(delta.begin(), delta.end(), 0) << std::endl;
}

#include <iostream>
#include <numeric>
#include <set>

int main() {
  std::multiset<int> l, r;
  int a, b;
  while (std::cin >> a >> b) {
    l.insert(a); r.insert(b);
  }
  std::cout << std::accumulate(l.begin(), l.end(), 0, [&r](int acc, int x) {
    return acc + x * r.count(x);
  }) << std::endl;
}

I expect much wailing and gnashing of teeth regarding the parsing, which of course is utterly trivial if you know a bit if regex.

I got bit by the input being more than one line. Embarrasing.

I wonder if any input starts with a "don't()" or if it's too early for Erik to pull such trickery.

[ inputs / "" ] | [.,.[0]|length] as [$H,$W] |

#----- In bound selectors -----#
def x: select(. >= 0 and . < $W);
def y: select(. >= 0 and . < $H);

reduce (
  [
    to_entries[] | .key as $y | .value |
    to_entries[] | .key as $x | .value |
    [ [$x,$y],. ]  | select(last!=".")
  ] | group_by(last)[] # Every antenna pair #
    | combinations(2)  | select(first < last)
) as [[[$ax,$ay]],[[$bx,$by]]] ({};
  # Assign linear anti-nodes #
  .[ range(-$H;$H) as $i | "\(
    [($ax+$i*($ax-$bx)|x), ($ay+$i*($ay-$by)|y)] | select(length==2)
  )"] = true
) | length

There's probably a smarter way to do this...

For part 1, I dumped everything into a matrix. Then I scanned it element by element. If I found an X, I searched in 8 directions from there and counted up if I found M A S in sequence.

For part 2 I searched for an A, checked each diagonal corner, and counted up if the opposites were M S or S M

https://github.com/gustafe/aoc2024/blob/main/d04-Ceres-Search.pl

"./input.actual"
|> System.IO.File.ReadAllText
|> fun source -> 
    System.Text.RegularExpressions.Regex.Matches(source, @"don't\(\)|do\(\)|mul\((\d+),(\d+)\)")
    |> Seq.fold
        (fun (acc, enabled) m ->
            match m.Value with
            | "don't()" -> acc, false
            | "do()" -> acc, true
            | mul when enabled && mul.StartsWith("mul") ->
                let (x, y) = int m.Groups.[1].Value, int m.Groups.[2].Value
                acc + x * y, enabled
            | _ -> acc, enabled ) 
        (0, true)
    |> fst
|> printfn "The sum of all valid multiplications with respect to do() and don't() is %A"

urgh this took me much longer than it should have... part 1 was easy enough but then I got tied up in knots with part 2. Finally I just sorto-bogo-sorted it all into shape

Perl: https://github.com/gustafe/aoc2024/blob/main/d05-Print-Queue.pl

Recheck array size: 98
All rechecks passed after 5938 passes
Duration: 00h00m00s (634.007 ms)

reduce (
  inputs |   scan("do\\(\\)|don't\\(\\)|mul\\(\\d+,\\d+\\)")
         | [[scan("(do(n't)?)")[0]], [ scan("\\d+") | tonumber]]
) as [[$do], [$a,$b]] (
  { do: true, s: 0 };
    if $do == "do" then .do = true
  elif $do         then .do = false
  elif .do         then .s = .s + $a * $b end
) | .s

RegExp(r"(?:don\'t\(\)(?:.(?<!do\(\)))*do\(\))|(?:mul\((\d*),(\d*)\))") .allMatches(input) .fold<int>( 0, (p, e) => p + (e.group(0)![0] != 'd' ? e.groups([1, 2]).fold<int>(1, (p, f) => p * int.parse(f!)) : 0))

*ecma balls

use strict;
use List::Util qw( min max );

open(FH, '<', $ARGV[0]) or die $!;

my @left;
my @right;

while (<FH>) {
	my @nums = split /\s+/, $_;
	push(@left, $nums[0]);
	push(@right, $nums[1]);
}

@left = sort { $b <=> $a } @left;
@right = sort { $b <=> $a } @right;

my $dist = 0;
my $sim = 0;
my $i = 0;

foreach my $lnum (@left) {
	$sim += $lnum * grep { $_ == $lnum } @right;

	my $rnum = $right[$i++];
	$dist += max($lnum, $rnum) - min($lnum, $rnum);
}

print 'Part 1: ', $dist, "\n";
print 'Part 2: ', $sim, "\n";

close(FH);

Except that gives you all the ways to reach 9s from a 0, which is part 2. For part 1, I changed the scores to be sets of reachable 9s, and the score of a square was the size of the set at that position.

#include <iostream>
#include <vector>
#include <algorithm>
#include <sstream>
#include <string>
#include <unordered_map>

std::unordered_multimap<int, int> ordering;

bool sorted_before(int a, int b) {
  auto range = ordering.equal_range(a);
  for (auto it = range.first; it != range.second; ++it) {
    if (it->second == b) return true;
  }
  return false;
}

int main() {
  int sum = 0;
  std::string line;
  while (std::getline(std::cin, line) && !line.empty()) {
    int l, r;
    char c;
    std::istringstream iss(line);
    iss >> l >> c >> r;
    ordering.insert(std::make_pair(l,r));
  }
  while (std::getline(std::cin, line)) {
    std::istringstream iss(line);
    std::vector<int> pages;
    int page;
    while (iss >> page) {
      pages.push_back(page);
      iss.get();
    }   
    std::vector<int> sorted_pages = pages;
    std::stable_sort(sorted_pages.begin(), sorted_pages.end(), sorted_before);
    if (pages == sorted_pages) {  // Change to != for 5-2
      sum += sorted_pages[sorted_pages.size()/2];
    }
  }
  std::cout << "Sum: " << sum << std::endl;
}

I treated the input as a list of strings and searched for "XMAS" and "SAMX", rotating and transposing the list to scan for vertical and diagonal words.

In my arrogance, I thought I could do the diagonal rotation easily, but I was very wrong. I got it out in the end, though.

Part 1. A quick look at the data showed that the input length was short enough to perform an O(2^n^) search with some early exits. I coded it as a dfs.

Part 2. Adding concatenation just changes the base from 2 to 3, which, while strictly slower, wasn't much slower for this input.

void d7(bool sub) => print(getLines()
    .map((l) => l.split(RegExp(r':? ')).map(int.parse).toList())
    .fold<int>(
        0, (p, ops) => test(ops, ops[0], ops[1], 2, sub) ? ops[0] + p : p));

bool test(List<int> l, int cal, int cur, int i, bool sub) =>
    cur == cal && i == l.length ||
    (i < l.length && cur <= cal) &&
        (test(l, cal, cur + l[i], i + 1, sub) ||
            test(l, cal, cur * l[i], i + 1, sub) ||
            (sub && test(l, cal, cur.concat(l[i]), i + 1, sub)));

I am pretty sure I did an OK job at avoiding edge cases, though I suspect this problem has many of them.

The complex part was the "look for a hole" part. My input size was 19999, meaning an O(n^2^) solution was probably fast enough, but I decided to optimise and use a min heap for each space size prematurely. I foresaw that you need to split up a space if it is oversized for a particular piece of data, i.e. pop the slot from the heap, reduce the size of the slot, and put it in the heap corresponding to the new size.

As it turns out, that instinct was also correct. By drawing the sample data (or counting or printing it out), I noticed that every page number had a defined relation with every other page number. This meant that a total ordering (rather than a lattice) existed, meaning it was possible to construct a comparison function.

So, the algorithm for part 1 was to check if a list was sorted, and part 2 was to sort the list. There's probably a 1-3 line solution for both parts a and b, but that's for Mr. The Reader.

Context: I participated (and completed!) in AoC last year and pragmatically wrote my code as a set of utility modules for solving these pathological problems. So, I had about 80% of the boilerplate for this problem written, waiting for me to read and relearn.

Anyway, the analysis: P1. was pretty straightforward. Just walk along the map, turn right if you hit an obstacle, and stop when you leave the map. I guessed that there may be a case where one needs to turn in place more than once to escape an obstacle, but I never checked if that was true. Either way, I got the answer out. This is a worst-case O(n^2^) solution, which was fast for n = 130.

P2. I chose to brute force this, and it was fine. I iterated through the grid, placing a wall if possible, and checked if this produced a loop using an explored set. This is worst case O(n^4^), which, for n=130, takes a few seconds to spit out the answer. It's parallelisable, though, so there's that. If a faster solution existed, I'd love to know.