day15.rs

//! # Timing is Everything
//!
//! Part one is the [Chinese Remainder Theorem](https://en.wikipedia.org/wiki/Chinese_remainder_theorem).
//! The integers n₁, n₂, ... nₖ map to the disc sizes which happen to be prime. This satisfies the
//! requirement that the integers are [pairwise coprime](https://en.wikipedia.org/wiki/Coprime_integers#Coprimality_in_sets).
//!
//! For simplicity we use the "search by sieving" method. We start at zero with a step the size of
//! the first integer. Then we search for each subsequent integer located at the correct offset of
//! minutes and multiply the step by the new integer. This preserves the relative offset at each
//! step in the next search.
use crate::util::iter::*;
use crate::util::parse::*;

type Pair = [usize; 2];

pub fn parse(input: &str) -> Pair {
    let disks: Vec<Pair> = input.iter_unsigned().skip(1).step_by(2).chunk::<2>().collect();
    let (part1, step) = solve(&disks, 0, 0, 1);
    let (part2, _step) = solve(&[[11, 0]], disks.len(), part1, step);
    [part1, part2]
}

pub fn part1(input: &Pair) -> usize {
    input[0]
}

pub fn part2(input: &Pair) -> usize {
    input[1]
}

fn solve(discs: &[Pair], offset: usize, mut time: usize, mut step: usize) -> (usize, usize) {
    for (o, &[size, position]) in discs.iter().enumerate() {
        while !(time + offset + o + 1 + position).is_multiple_of(size) {
            time += step;
        }
        step *= size;
    }

    (time, step)
}