Day 9 (Advent of Code 2020)

Contents

• Part 2

Day 9's problem statement is convoluted - the "ah maybe that's why I don't usually do Advent of Code" kind of convoluted, but let's give it a go anyway.

So, we have a series of numbers, like so:

35
20
15
25
47
40
62
55
65
95
102
117
150
182
127
219
299
277
309
576

And uh the first N numbers are a "preamble" and every number that comes after that must be the sum of any two of the numbers that come before it.

For the example above, N is 5. So, again, there's probably a smart and fast way to solve this, but I'll go again for a simple and correct solution instead.

One thing I like about this problem is that it lets me showcase a bunch of cool methods.

We're going to iterate over windows of size n+1 - so here, elements 0..=5 (inclusive), then 1..=6, then 2..=7, and so on. Then we're going to get all possible combinations of elements 0..5 (exclusive), 1..6, 2..7, and see if the sum of any of those combinations is equal to the last element of our window.

use itertools::Itertools;

fn main() {
    let numbers = include_str!("input.txt")
        .lines()
        .map(|x| x.parse::<usize>().unwrap())
        .collect::<Vec<_>>();

    let n = 5;
    let answer = numbers.windows(n + 1).find_map(|s| {
        if (&s[..n])
            .iter()
            .tuple_combinations()
            .any(|(a, b)| a + b == s[n])
        {
            None
        } else {
            Some(s[n])
        }
    });
    println!("answer = {:?}", answer);
}

$ cargo run --quiet
answer = Some(127)

Let's try it with n = 25 and the actual input:

$ cargo run --quiet
answer = Some(26134589)

Hey, that's the correct answer!

Part 2

The next part asks us to find "a contiguous set of at least two numbers in our list which sum to the invalid number from step 1".

Well, that doesn't seem too hard either. One thing we can do is to sum all the windows of size 2, then all the windows of size 3, and so on - and sum the items in all of these. As soon as we reach the answer, we're done!

    let answer = answer.unwrap();

    let answer2 = (2..numbers.len())
        .into_iter()
        .map(|n| numbers.windows(n).map(|s| s.iter().sum::<usize>()))
        .flatten()
        .find(|&n| n == answer);
    println!("answer2 = {:?}", answer2);

$ cargo run --quiet
answer2 = Some(26134589)

Ok, so we did find a contiguous set of numbers whose sum is the same as the answer we found in part 1, but we don't know where or how large the set was.

Let's address that:

    let answer2 = (2..numbers.len())
        .into_iter()
        .map(|n| {
            numbers
                .windows(n)
                .enumerate()
                .map(move |(i, s)| (n, i, s.iter().sum::<usize>()))
        })
        .flatten()
        .find(|&(_, _, sum)| sum == answer);

    let (n, i, _) = answer2.unwrap();
    let set = &numbers[i..][..n];
    println!("sum({:?}) = {}", set, answer);

$ cargo run --quiet
sum([1503494, 978652, 1057251, 1142009, 1239468, 1407633, 1048040, 1484541, 1164289, 1432864, 1792914, 2556472, 2464510, 1750429, 1753116, 1673488, 1685419]) = 26134589

That's better.

Now we need to add together the smallest and largest number in this contiguous range:

    let answer3 = set.iter().max().unwrap() + set.iter().min().unwrap();
    dbg!(answer3);

$ cargo run --quiet
[src/main.rs:39] answer3 = 3535124

Aaand we're done! That was easy, I don't know what I was worried about.