Function aoc::y15d15::y15d15

pub fn y15d15(input: &str, part: u32) -> i32

The solution for the day fifteen challenge.

As usual we take the input as a string and a variable to determine which part we’re computing (in part 2 we need to limit the options to cookies that result in 500 calories). We start by processing the input and building Ingredients to keep track of the various qualities of the ingredients. We then create a vector of vectors, one for each ingredient where the sub-vector are all of the possible teaspoons of the ingredient to use to make the cookie (0 to 100). We then take the cartesian product of those vectors to get all of the possible recipes (both valid and invalid) and then filter the result to only the valid options (i.e., that use 100 total teaspoons, and in part 2 are 500 calories). With the list of valid recipes we compute the score of each one and add it to our std::collections::BinaryHeap. To get the highest possible score we just pop the heap.

Example

// probably read this from the input file...
let input = concat!(
    "Candy: capacity -1, durability 3, flavor 1, texture 1, calories 2\n",
    "Chocolate: capacity 2, durability 2, flavor 1, texture 1, calories 8",
);
assert_eq!(y15d15(input, 1), 400000000);
assert_eq!(y15d15(input, 2), 125000000);