Updated version using my own ETS cache, much quicker which is interesting . . .
Solution for 2024 day 11
part_one: **193607** in **3.71ms**
part_two: **229557103025807** in **59.43ms**
defmodule Aoc2024.Solutions.Y24.Day11 do
require Integer
alias AoC.Input
def parse(input, _part) do
Input.read!(input)
|> String.trim()
|> String.split(" ", trim: true)
|> Enum.map(&String.to_integer/1)
end
def part_one(problem) do
:ets.new(:apply_rules_cache, [:named_table, :public, :set, :protected])
result =
Enum.reduce(problem, 0, fn stone, acc ->
acc + apply_rules(stone, 0, 25)
end)
:ets.delete(:apply_rules_cache)
result
end
def part_two(problem) do
:ets.new(:apply_rules_cache, [:named_table, :public, :set, :protected])
result =
Enum.reduce(problem, 0, fn stone, acc ->
acc + apply_rules(stone, 0, 75)
end)
:ets.delete(:apply_rules_cache)
result
end
def apply_rules(stone, count, target) do
case :ets.lookup(:apply_rules_cache, {stone, count, target}) do
[{_, result}] ->
result
[] ->
result =
if count == target do
1
else
target_stones = rule(stone)
case target_stones do
[_first, _second] ->
Enum.reduce(target_stones, 0, fn stone, acc ->
acc + apply_rules(stone, count + 1, target)
end)
stone ->
apply_rules(stone, count + 1, target)
end
end
:ets.insert(:apply_rules_cache, {{stone, count, target}, result})
result
end
end
def rule(stone) do
if stone == 0 do
1
else
stone_length = Integer.digits(stone)
num_digits = length(stone_length)
# stone_string = Integer.to_string(stone)
# digits = byte_size(stone_string)
case Integer.is_even(num_digits) do
true ->
stone_string = Integer.to_string(stone)
mid = div(num_digits, 2)
<<left::binary-size(mid), right::binary>> = stone_string
[String.to_integer(left), String.to_integer(right)]
_ ->
stone * 2024
end
end
end
end
