1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84
//! # Syntax Scoring
//!
//! This day is a variation of the classic parentheses balancing problem. To solve we use a `vec`
//! as a stack, pushing opening delimiters onto the stack, then popping the top of the stack
//! whenever we encounter a closing delimiter. If there is a mismatch between opening and closing
//! delimiters then we return the specified error value immediately.
//!
//! For part 2 the completion score is the remaining items on the stack, reversed and converted from
//! corresponding closing delimiters. For example the completion string `])}>` would have a stack
//! that looks like `<{([`, where the right hand side is the top of the stack.
pub fn parse(input: &str) -> Vec<&[u8]> {
input.lines().map(str::as_bytes).collect()
}
pub fn part1(input: &[&[u8]]) -> u64 {
let mut stack = Vec::new();
let mut score = 0;
for line in input {
score += syntax_score(line, &mut stack);
stack.clear();
}
score
}
pub fn part2(input: &[&[u8]]) -> u64 {
let mut stack = Vec::new();
let mut scores = Vec::new();
for line in input {
if syntax_score(line, &mut stack) == 0 {
scores.push(autocomplete_score(&stack));
}
stack.clear();
}
scores.sort_unstable();
scores[scores.len() / 2]
}
fn syntax_score(line: &[u8], stack: &mut Vec<u8>) -> u64 {
for &b in line {
match b {
b'(' | b'[' | b'{' | b'<' => stack.push(b),
b')' => {
if stack.pop().unwrap() != b'(' {
return 3;
}
}
b']' => {
if stack.pop().unwrap() != b'[' {
return 57;
}
}
b'}' => {
if stack.pop().unwrap() != b'{' {
return 1197;
}
}
b'>' => {
if stack.pop().unwrap() != b'<' {
return 25137;
}
}
_ => unreachable!(),
}
}
0
}
fn autocomplete_score(stack: &[u8]) -> u64 {
fn helper(b: u8) -> u64 {
match b {
b'(' => 1,
b'[' => 2,
b'{' => 3,
b'<' => 4,
_ => unreachable!(),
}
}
stack.iter().rev().fold(0, |acc, &b| 5 * acc + helper(b))
}