Generate them!
mudasobwa
Creator of Cure
marcelfahle
where are the reduce_while folks? 
defmodule ExAOC2022.Day6 do
@input "./lib/day6_input.txt"
def puzzle1() do
@input
|> File.read!()
|> distinct_chars_idx(4)
end
def puzzle2() do
@input
|> File.read!()
|> distinct_chars_idx(14)
end
defp distinct_chars_idx(str, len) do
str
|> String.graphemes()
|> Enum.with_index()
|> Enum.reduce_while([], fn {char, idx}, acc ->
if length(acc) < len do
{:cont, [char | acc]}
else
if has_dups?(acc) do
{:cont, [char | shorten(acc)]}
else
{:halt, idx}
end
end
end)
end
defp has_dups?(list), do: Enum.uniq(list) != list
defp shorten(list), do: list |> Enum.reverse() |> tl() |> Enum.reverse()
end
0
newton-peixoto
Could you give an example how to do it? I believe you`ll use macros but I dont know much about it would be nice if you could talk a little about it
0
adamu
dams
Very short code, it’s basically 3 lines 
because there isn’t much complexity. Forced myself to use pattern matching for fun.
defmodule Day6 do
# patternmatching fun!
# l = the list of chars, p = position where we are in the list
# s = the size of consecutive non-repetitive chars we're looking for
def find_marker(l, p, s ), do: find_marker(l, p, s, length(Enum.uniq(Enum.take(l,s))))
def find_marker(_, p, s, s), do: p+s
def find_marker(l, p, s, _), do: find_marker(tl(l), p+1, s)
end
list = File.read!("input") |> String.to_charlist()
Day6.find_marker(list, 0, 4) |> IO.inspect() # part 1
Day6.find_marker(list, 0, 14) |> IO.inspect() # part 2
1
deadbeef
Virtually identical to others
https://github.com/ed-flanagan/advent-of-code-solutions-elixir/blob/main/lib/advent/y2022/d06.ex
1
bmitc
Today’s was weird. Part two didn’t require a single change aside from the window size. Just a single function call of:
@doc """
Finds the index of the next element after the first window of the given size
that contains unique elements
"""
@spec find_end_of_first_distinct_window([any()], pos_integer()) :: pos_integer()
def find_end_of_first_distinct_window(data_stream, window_size) do
data_stream
|> Enum.chunk_every(window_size, 1)
|> Enum.find_index(&Utilities.unique_elements?/1)
|> Kernel.+(window_size)
end
This pipeline is literally just the problem statement. 
The function Utilities.unique_elements? just compares the length of an enum to the length of Enum.uniq applied to the same enum. Full Livebook notebook here.
0
LostKobrakai
That’s only because elixir has very expressive means of working with windows on top of enumerables. If you consider imperative languages then var a; var b; var c; var d is not as simply replaced with 14 manually juggled variables.
1
tfwright
I should have reread the docs of chunk_every once more before deciding it wouldn’t work for this. I manually accumulated a list like a chump 
0
akaibukai
Very very clean recursive solution! I like it!
Edit: FWIW, you can also do that:
Day6.find_marker(list, 0, 4) |> IO.inspect(label: "Part 1")
Day6.find_marker(list, 0, 14) |> IO.inspect(label: "Part 2")
1