use crate::prelude::*;

pub fn solution() -> Solution {
  Solution::new(Day::new(13, 2020), part_1, part_2)
    .with_expected(1895, 840493039281088_i64)
}

fn part_1(input: &str) -> Result<String> {
  let mut lines = input.lines();

  let target = lines.next().map(str::parse::<i32>).unwrap().unwrap();
  let busses = lines
    .next()
    .unwrap()
    .split(',')
    .filter_map(|value| value.parse().ok())
    .collect::<Vec<i32>>();

  let mut difference = 0;
  let mut earliest_bus = 0;
  let mut minutes = 0;

  for bus in busses {
    let rest = target % bus;

    if minutes < rest {
      difference = bus - rest;
      earliest_bus = bus;
      minutes = rest;
    }
  }

  Ok((earliest_bus * difference).to_string())
}

fn part_2(input: &str) -> Result<String> {
  // Skip the first line, as it isn't used for the second part of the puzzle.
  let input = input.lines().nth(1).unwrap();

  let busses = input
    .split(',')
    .enumerate()
    .filter_map(|(offset, bus)| {
      if let Ok(bus) = bus.parse() {
        Some((offset, bus))
      } else {
        None
      }
    })
    .collect::<Vec<(usize, usize)>>();

  let mut timestamp = 0;
  // Start the increment at 1 as we'll be multiplying it with the bus numbers.
  let mut increment = 1;
  // Keep track of which busses we've gotten modulo equals 0 from.
  let mut used_busses = vec![];

  'outer: loop {
    timestamp += increment;

    for (offset, bus) in &busses {
      if (timestamp + offset) % bus != 0 {
        continue 'outer;
      } else if !used_busses.contains(&(offset, bus)) {
        used_busses.push((offset, bus));
        increment *= bus;
      }
    }

    break 'outer;
  }

  Ok(timestamp.to_string())
}