# Day 9 (Advent of Code 2020)

This article is part of the Advent of Code 2020 series.

- Part 2

### Contents

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.

useitertools::Itertools;fnmain(){letnumbers =include_str!("input.txt").lines().map(|x| x.parse::<usize>().unwrap()).collect::<Vec<_>>();letn =5;letanswer = 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)

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

Let's try it with `n = 25`

and the actual input:

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

Hey, that's the correct answer!

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!

letanswer = answer.unwrap();letanswer2 =(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:

letanswer2 =(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();letset =&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:

letanswer3 = 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.

This article is part 9 of the Advent of Code 2020 series.

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