solve and optimize year 2022 day 03

src/year2022/day03.rs

@@ -1,31 +1,39 @@
-use std::collections::BTreeSet;
-
-fn char_to_score(c: char) -> u32 {
-    match c {
-        'a'..='z' => c as u32 - 96,
-        'A'..='Z' => c as u32 - 38,
-        _ => unreachable!(),
+/// Number of trailing zeros tells us which bit is set and there should only be one set
+/// Do some math to convert to the needed priority
+fn bits_to_priority(b: u64) -> u32 {
+    let trailing = b.trailing_zeros();
+    match trailing {
+        0..=25 => trailing + 27,
+        32..=57 => trailing - 31,
+        _ => unreachable!(""),
     }
 }
 
+/// For each byte in the string offset it towards 0
+/// to make it fit in u64 and set a bit at the position of its value
+fn to_bits(s: &str) -> u64 {
+    s.bytes().fold(0, |a, v| a | 1 << (v - 65))
+}
+
 pub fn part1(input: &str) -> impl std::fmt::Display {
     input
         .lines()
         .map(|l| l.split_at(l.len() / 2))
-        .map(|(l, r)| {
-            let a: BTreeSet<char> = l.chars().collect();
-            let b: BTreeSet<char> = r.chars().collect();
-            a.intersection(&b)
-                .next()
-                .expect("We know there is at least one")
-                .clone()
-        })
-        .map(char_to_score)
+        .map(|(l, r)| bits_to_priority(to_bits(l) & to_bits(r)))
         .sum::<u32>()
 }
 
 pub fn part2(input: &str) -> impl std::fmt::Display {
-    ""
+    let mut lines = input.lines();
+    let mut sum = 0;
+
+    while let Some(l1) = lines.next() {
+        let l2 = lines.next().expect("second is guaranteed to be there");
+        let l3 = lines.next().expect("third is guaranteed to be there");
+
+        sum += bits_to_priority(to_bits(l1) & to_bits(l2) & to_bits(l3));
+    }
+    sum
 }
 
 #[test]