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Dec 22, 2020 clock icon 4 min #rust · #advent-of-code

Day 9 (Advent of Code 2020)

From the series Advent of Code 2020
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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.

Rust code
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);
}
Shell session
$ cargo run --quiet
answer = Some(127)
Bear

Cool, this matches the example! I guess we're already done?

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

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

Hey, that's the correct answer!

Bear

Onwards!

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!

Rust code
    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);
Shell session
$ 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:

Rust code
    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);
Shell session
$ 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:

Rust code
    let answer3 = set.iter().max().unwrap() + set.iter().min().unwrap();
    dbg!(answer3);
Shell session
$ 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.

You're reading the Advent of Code 2020 series.
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