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submitted 8 months ago* (last edited 8 months ago) by Ategon@programming.dev to c/advent_of_code@programming.dev

Day 9: Mirage Maintenance

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[-] capitalpb@programming.dev 1 points 8 months ago

A pretty simple one today, but fun to do. I could probably clean up the parsing code (AKA my theme for this year), and create just one single vector instead of having the original history separated out from all of the sequences, but this is what made sense to me on my first pass so it's how I did it.

https://github.com/capitalpb/advent_of_code_2023/blob/main/src/solvers/day09.rs

pub struct Day09;

fn get_history(input: &str) -> Vec {
    input
        .split(' ')
        .filter_map(|num| num.parse::().ok())
        .collect::>()
}

fn get_sequences(history: &Vec) -> Vec> {
    let mut sequences = vec![get_steps(&history)];

    while !sequences.last().unwrap().iter().all_equal() {
        sequences.push(get_steps(sequences.last().unwrap()));
    }

    sequences
}

fn get_steps(sequence: &Vec) -> Vec {
    sequence
        .iter()
        .tuple_windows()
        .map(|(x, y)| y - x)
        .collect()
}

impl Solver for Day09 {
    fn star_one(&self, input: &str) -> String {
        input
            .lines()
            .map(|line| {
                let history = get_history(line);

                let add_value = get_sequences(&history)
                    .iter()
                    .rev()
                    .map(|seq| seq.last().unwrap().clone())
                    .reduce(|acc, x| acc + x)
                    .unwrap();

                history.last().unwrap() + add_value
            })
            .sum::()
            .to_string()
    }

    fn star_two(&self, input: &str) -> String {
        input
            .lines()
            .map(|line| {
                let history = get_history(line);

                let minus_value = get_sequences(&history)
                    .iter()
                    .rev()
                    .map(|seq| seq.first().unwrap().clone())
                    .reduce(|acc, x| x - acc)
                    .unwrap();

                history.first().unwrap() - minus_value
            })
            .sum::()
            .to_string()
    }
}
this post was submitted on 09 Dec 2023
20 points (95.5% liked)

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