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

Elixir

defmodule AdventOfCode.Solution.Year2024.Day02 do
  use AdventOfCode.Solution.SharedParse

  @impl true
  def parse(input) do
    for line <- String.split(input, "\n", trim: true),
        do: String.split(line) |> Enum.map(&String.to_integer/1)
  end

  def part1(input) do
    Enum.count(input, &is_safe(&1, false))
  end

  def part2(input) do
    Enum.count(input, &(is_safe(&1, true) or is_safe(tl(&1), false)))
  end

  def is_safe([a, b, c | rest], can_fix) do
    cond do
      (b - a) * (c - b) > 0 and abs(b - a) in 1..3 and abs(c - b) in 1..3 ->
        is_safe([b, c | rest], can_fix)

      can_fix ->
        is_safe([a, c | rest], false) or is_safe([a, b | rest], false)

      true ->
        false
    end
  end

  def is_safe(_, _), do: true
end

Elixir

defmodule AdventOfCode.Solution.Year2024.Day03 do
  def part1(input) do
    Regex.scan(~r/mul\((?<l>\d+),(?<r>\d+)\)/, input, capture: ["l", "r"])
    |> Stream.map(fn l -> Enum.reduce(l, 1, &(&2 * String.to_integer(&1))) end)
    |> Enum.sum()
  end

  def part2(input) do
    input |> String.replace(~r/don't\(\).*(do\(\)|$)/Us, "") |> part1
  end
end

Elixir

defmodule AdventOfCode.Solution.Year2024.Day04 do
  use AdventOfCode.Solution.SharedParse

  defmodule Map do
    defstruct [:chars, :width, :height]
  end

  @impl true
  def parse(input) do
    chars = String.split(input, "\n", trim: true) |> Enum.map(&String.codepoints/1)
    %Map{chars: chars, width: length(Enum.at(chars, 0)), height: length(chars)}
  end

  def at(%Map{} = map, x, y) do
    cond do
      x < 0 or x >= map.width or y < 0 or y >= map.height -> ""
      true -> map.chars |> Enum.at(y, []) |> Enum.at(x, "")
    end
  end

  def part1(map) do
    dirs = for dx <- -1..1, dy <- -1..1, {dx, dy} != {0, 0}, do: {dx, dy}
    xmas = String.codepoints("XMAS") |> Enum.with_index() |> Enum.drop(1)

    for x <- 0..(map.width - 1),
        y <- 0..(map.height - 1),
        "X" == at(map, x, y),
        {dx, dy} <- dirs,
        xmas
        |> Enum.all?(fn {c, n} -> at(map, x + dx * n, y + dy * n) == c end),
        reduce: 0 do
      t -> t + 1
    end
  end

  def part2(map) do
    for x <- 0..(map.width - 1),
        y <- 0..(map.height - 1),
        "A" == at(map, x, y),
        (at(map, x - 1, y - 1) <> at(map, x + 1, y + 1)) in ["MS", "SM"],
        (at(map, x - 1, y + 1) <> at(map, x + 1, y - 1)) in ["MS", "SM"],
        reduce: 0 do
      t -> t + 1
    end
  end
end

Elixir

defmodule AdventOfCode.Solution.Year2024.Day05 do
  use AdventOfCode.Solution.SharedParse

  @impl true
  def parse(input) do
    [rules, pages_list] =
      String.split(input, "\n\n", limit: 2) |> Enum.map(&String.split(&1, "\n", trim: true))

    {for(rule <- rules, do: String.split(rule, "|") |> Enum.map(&String.to_integer/1))
     |> MapSet.new(),
     for(pages <- pages_list, do: String.split(pages, ",") |> Enum.map(&String.to_integer/1))}
  end

  def part1({rules, pages_list}), do: solve(rules, pages_list, false)

  def part2({rules, pages_list}), do: solve(rules, pages_list, true)

  def solve(rules, pages_list, negate) do
    for pages <- pages_list, reduce: 0 do
      total ->
        ordered = Enum.sort(pages, &([&1, &2] in rules))

        if negate != (ordered == pages),
          do: total + Enum.at(ordered, div(length(ordered), 2)),
          else: total
    end
  end
end

You can run it on Linux through Firejail.

Well, if you use a password manager such as bitwarden you can store your 2FA one ctrl-v away. Even if this is a less secure setup, that still prevents someone eavesdropping on your password from reusing it.

Why would you not want to use 2FA?

I’m also using it in blind mode save for the symbols. I really like it so far.

And I got a reply: “we are working on it”. This is great news, let’s see what hapens!

I’ve sent them a main as well this morning. The dictionaries apk are also only available using http.

I cannot seem to find either the game app or the dictionaries on the play store.

It looks really neat, but is it still maintained?

Of course. I was referring to the non-self-hosted solution.

My bad indeed, I thought Lemmy supported OAuth but I was confused with Mastodon. Hopefully someone contributes a OAuth/OIDC solution soon.

Why do you ask for the user password rather than using oauth to access the user account? This looks highly suspicious, websites should never do that.

Interesting, but it lacks major backend parts such as instruction selection and register allocation.