aoc/year2018/
day01.rs

1//! # Chronal Calibration
2//!
3//! The simplest approach to part two is to store previously seen numbers in a `HashSet` then
4//! stop once a duplicate is found. However this approach requires scanning the input of ~1,000
5//! numbers multiple times, around 150 times for my input.
6//!
7//! A much faster `O(nlogn)` approach relies on the fact that each frequency increases by the same
8//! amount (the sum of all deltas) each time the list of numbers is processed. For example:
9//!
10//! ```none
11//!    Deltas: +1, -2, +3, +1 =>
12//!    0    1   -1    2
13//!    3    4    2    5
14//! ```
15//!
16//! Two frequencies that are a multiple of the sum will eventually repeat. First we group each
17//! frequencies by its remainder modulo the sum, using `rem_euclid` to handle negative frequencies
18//! correctly, Then we sort, first by the remainder to group frequencies that can repeat together,
19//! then by the frequency increasing in order to help find the smallest gap between similar
20//! frequencies, then lastly by index as this is needed in the next step.
21//!
22//! For the example this produces `[(0, 0, 0), (1, 1, 1), (2, -1, 2), (2, 2, 3)]`. Then we use
23//! a sliding windows of size two to compare each pair of adjacent canditates, considering only
24//! candidates with the same remainder. For each valid pair we then produce a tuple of
25//! `(frequency gap, index, frequency)`.
26//!
27//! Finally we sort the tuples in ascending order, first by smallest frequency gap, breaking any
28//! ties using the index to find frequencies that appear earlier in the list. The first tuple
29//! in the list gives the result, in the example this is `[(3, 2, 2)]`.
30use crate::util::parse::*;
31
32pub fn parse(input: &str) -> Vec<i32> {
33    input.iter_signed().collect()
34}
35
36pub fn part1(input: &[i32]) -> i32 {
37    input.iter().sum()
38}
39
40pub fn part2(input: &[i32]) -> i32 {
41    // The frequencies increase by this amount each pass through the list of deltas.
42    let total: i32 = input.iter().sum();
43
44    // Calculate tuples of `(frequency gap, index, frequency)` then sort to group frequencies that
45    // can collide together.
46    let mut frequency: i32 = 0;
47    let mut seen = Vec::with_capacity(input.len());
48
49    for n in input {
50        seen.push((frequency.rem_euclid(total), frequency, seen.len()));
51        frequency += n;
52    }
53
54    seen.sort_unstable();
55
56    // Compare each adjacent pair of tuples to find candidates, then sort by smallest gap first,
57    // tie breaking with index if needed.
58    let mut pairs = Vec::new();
59
60    for window in seen.windows(2) {
61        let (remainder0, freq0, index0) = window[0];
62        let (remainder1, freq1, _) = window[1];
63
64        if remainder0 == remainder1 {
65            pairs.push((freq1 - freq0, index0, freq1));
66        }
67    }
68
69    pairs.sort_unstable();
70
71    // Result is the frequency of the first tuple.
72    let (_, _, freq) = pairs[0];
73    freq
74}
aoc/year2018/day01.rs

aoc/year2018/
day01.rs
