# Advent of Code 2023 - Day 2

Ready for Day 2? Let’s jump right into it.

You can find the final code @ advent-of-code/2023/day2

We have an input that represents a game, each game has an id and a set of
*hands*. Each hand has the number of cubes grabbed by the bag and its color.

Here’s the input example:

```
Game 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green
Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue
Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red
Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red
Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green
```

The first line represents game #1 that has 3 different hands.

We are given the number of cubes that are present in the bag and we have to determine which game is possible with that information. I’m going to quote the problem description to make things clearer

Which games would have been possible if the bag contained only 12 red cubes, 13 green cubes, and 14 blue cubes?

In the example above, games 1, 2, and 5 would have been possible if the bag had been loaded with that configuration. However, game 3 would have been impossible because at one point the Elf showed you 20 red cubes at once; similarly, game 4 would also have been impossible because the Elf showed you 15 blue cubes at once. If you add up the IDs of the games that would have been possible, you get 8.

In this case it could be very useful to create a struct that encapsulates all the game information that we find on each line. Each game has an id and its sets representation. Each set can be represented as a matrix of RGB values.

```
type RGB = (u32, u32, u32);
#[derive(Debug)]
struct Game {
id: u32,
sets: Vec<RGB>,
}
```

Now we can implement a `From<&str>`

trait for the `Game`

type so that we can use
it later to parse the game information from a single line. The parsing logic is
going to be a bit verbose but I will basically split by `:`

, `;`

and `,`

until I
only have RGB values that are going to be stored in the matrix.

```
impl From<&str> for Game {
fn from(value: &str) -> Self {
let (game_str, sets_str) = value.split_once(":").unwrap();
let id = game_str.split_once(" ").unwrap().1.parse().unwrap();
let mut sets = Vec::new();
for set in sets_str.split(";") {
let mut rgb = (0, 0, 0);
for hand in set.split(",") {
match hand.trim().split_once(" ").unwrap() {
(id, "red") => {
rgb.0 = id.parse().unwrap();
}
(id, "green") => {
rgb.1 = id.parse().unwrap();
}
(id, "blue") => {
rgb.2 = id.parse().unwrap();
}
_ => {}
}
}
sets.push(rgb);
}
Game { id, sets }
}
}
```

Quick test to validate our parsing implementation

```
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_game_parse() {
let g = Game::from("Game 123: 2 green, 1 red; 3 red, 4 blue");
assert_eq!(g.id, 123);
assert_eq!(g.sets, vec![(1, 2, 0), (3, 0, 4)]);
}
}
```

Everything looks good so far, with that let’s move on to the solution of the first part of the problem.

We have to determine if a game is possible given an initial cube configuration. A game is possible if the max number of cubes of a specific color grabbed in a game is less than or equal to the value in the configration.

To make things easier I would like to implement a method that returns the max value of a particular cube color grabbed in a game, so that we can make an immediate comparison when we iterate through all the games in the input.

```
impl Game {
fn get_max_rgb_value(&self) -> RGB {
let mut max_rgb = (0, 0, 0);
for (r, g, b) in &self.sets {
if max_rgb.0 < *r {
max_rgb.0 = *r;
}
if max_rgb.1 < *g {
max_rgb.1 = *g;
}
if max_rgb.2 < *b {
max_rgb.2 = *b;
}
}
max_rgb
}
}
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_max_rgb_value() {
let g1 = Game::from("Game 123: 2 green, 1 red; 3 red, 4 blue");
assert_eq!(g1.get_max_rgb_value(), (3, 2, 4));
let g2 = Game::from("Game 123: 1 red; 3 red, 4 blue");
assert_eq!(g2.get_max_rgb_value(), (3, 0, 4));
let g3 = Game::from("Game 123: 2 green");
assert_eq!(g3.get_max_rgb_value(), (0, 2, 0));
}
}
```

At this point we just have to iterate through all the games, once they’ve all been parsed, and sum the id of those that satisfy the configuration condition.

```
fn main() -> io::Result<()> {
let mut input = String::new();
io::stdin().read_to_string(&mut input)?;
writeln!(io::stdout(), "{}", part1(&input)?)?;
Ok(())
}
fn part1(input: &str) -> io::Result<u32> {
let config: RGB = (12, 13, 14);
let sum = input
.lines()
.map(Game::from)
.filter(|x| {
let rgb = x.get_max_rgb_value();
rgb.0 <= config.0 && rgb.1 <= config.1 && rgb.2 <= config.2
})
.map(|x| x.id)
.sum();
Ok(sum)
}
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_part1() {
assert_eq!(0, part1("Game 3: 13 red").unwrap());
assert_eq!(0, part1("Game 3: 14 green").unwrap());
assert_eq!(0, part1("Game 3: 15 blue").unwrap());
assert_eq!(1, part1("Game 1: 10 green; 5 blue").unwrap());
assert_eq!(1, part1("Game 1: 10 green; 5 blue").unwrap());
assert_eq!(1, part1("Game 1: 10 green; 5 blue").unwrap());
assert_eq!(2, part1("Game 2: 12 red").unwrap());
assert_eq!(2, part1("Game 2: 13 green").unwrap());
assert_eq!(2, part1("Game 2: 14 blue").unwrap());
}
}
```

`cat input | cargo run -`

returns the correct answer, let’s move to part 2 now.

The problem is now asking to calculate which is the minimum number of cubes and
their colors that could have made the game possible. If you followed along, you
may have noticed that we don’t need to code anymore logic for this. Indeed,
`get_max_rgb_value`

is all we need to answer that question since that
already returns what the problem is asking. Once we have the minimum number
of cubes that could have made the game possible, we have to multiply those
RGB values and sum all of them to get the final result.

```
fn part2(input: &str) -> io::Result<u32> {
let sum = input
.lines()
.map(Game::from)
.map(|x| x.get_max_rgb_value())
.map(|(r, g, b)| r * g * b)
.sum();
Ok(sum)
}
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn test_part2() {
assert_eq!(
2286,
part2(
r"Game 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green
Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue
Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red
Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red
Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green"
)
.unwrap()
);
}
}
```

Again, `cat input | cargo run -`

returns the correct solution.

I’ve been lucky this time around, day 2 is off the map and we can call it a day, yay!