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Simplified code for day 04

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This commit is contained in:
Dreaded_X 2023-12-04 11:47:29 +01:00
parent 3720115175
commit 76e36d43c1
Signed by: Dreaded_X
GPG Key ID: 5A0CBFE3C3377FAA
1 changed files with 26 additions and 32 deletions
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58
2023/src/bin/day04.rs
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@ -91,44 +91,38 @@ impl aoc::Solver for Day {
}
fn part2(input: &str) -> Self::Output2 {
// Calculate how many wins we have per game
let wins: Vec<_> = input
.lines()
.map(|line| {
// Get rid of the first part
let (_, line) = line
.split_once(": ")
.expect("Input should be formatted properly");
// Seperate the winning numbers and numbers we have
let (winning, numbers) = line
.split_once(" | ")
.expect("Input should be formatted properly");
// Parse the winning numbers
let winning: Vec<_> = winning.split(' ').flat_map(str::parse::<usize>).collect();
// Parse the numbers we have, check if they are a winning number and count the
// amount of winning numbers we have
numbers
.split(' ')
.flat_map(str::parse::<usize>)
.filter(|num| winning.contains(num))
.count()
})
.collect();
// Start out with one copy of every card
let mut copies = vec![1; wins.len()];
let mut copies = vec![1; input.lines().count()];
// Keep track of how many copies we get of every card
for (i, win) in wins.iter().enumerate() {
for j in (i + 1)..(i + win + 1) {
// Add the current amount of copies to the card we win
input.lines().enumerate().for_each(|(i, line)| {
// Get rid of the first part
let (_, line) = line
.split_once(": ")
.expect("Input should be formatted properly");
// Seperate the winning numbers and numbers we have
let (winning, numbers) = line
.split_once(" | ")
.expect("Input should be formatted properly");
// Parse the winning numbers
let winning: Vec<_> = winning.split(' ').flat_map(str::parse::<usize>).collect();
// Parse the numbers we have, check if they are a winning number and count the
// amount of winning numbers we have
let wins = numbers
.split(' ')
.flat_map(str::parse::<usize>)
.filter(|num| winning.contains(num))
.count();
// Increment the amount that we have of the cards that we win by the number we have of
// the current card
for j in (i + 1)..(i + wins + 1) {
// e.g. we have 2 copies of the current card that has 3 wins, this means we add 2
// to the next 3 cards
copies[j] += copies[i]
}
}
});
copies.iter().sum()
}