aoc/year2020/
day02.rs

1//! # Password Philosophy
2//!
3//! Parsing the rules upfront allows both part one and part two to be solved in a straightforward manner.
4//!
5//! There's no need to first convert the input into lines since we know that each rule has 4 parts.
6//! Instead we use the [`split`] method with a slice of delimiters to break the input into
7//! an `Iterator` of tokens, then use our utility [`chunk`] method to group the tokens into an
8//! array of size 4.
9//!
10//! [`split`]: slice::split
11//! [`chunk`]: crate::util::iter
12use crate::util::iter::*;
13use crate::util::parse::*;
14
15pub struct Rule<'a> {
16    start: usize,
17    end: usize,
18    letter: u8,
19    password: &'a [u8],
20}
21
22impl Rule<'_> {
23    fn from([a, b, c, d]: [&str; 4]) -> Rule<'_> {
24        Rule {
25            start: a.unsigned(),
26            end: b.unsigned(),
27            letter: c.as_bytes()[0],
28            password: d.as_bytes(),
29        }
30    }
31}
32
33pub fn parse(input: &str) -> Vec<Rule<'_>> {
34    input
35        .split(['-', ':', ' ', '\n'])
36        .filter(|s| !s.is_empty())
37        .chunk::<4>()
38        .map(Rule::from)
39        .collect()
40}
41
42pub fn part1(input: &[Rule<'_>]) -> usize {
43    input
44        .iter()
45        .filter(|rule| {
46            let count = rule.password.iter().filter(|&&l| l == rule.letter).count();
47            (rule.start..=rule.end).contains(&count)
48        })
49        .count()
50}
51
52pub fn part2(input: &[Rule<'_>]) -> usize {
53    input
54        .iter()
55        .filter(|rule| {
56            let first = rule.password[rule.start - 1] == rule.letter;
57            let second = rule.password[rule.end - 1] == rule.letter;
58            first ^ second
59        })
60        .count()
61}
aoc/year2020/day02.rs

aoc/year2020/
day02.rs
