Solved AOC 2024 day 4

2024/src/day04.rs

@@ -0,0 +1,139 @@
+const INPUT: &str = include_str!("../../2024/input/day04.txt");
+
+pub(crate) fn part1() -> usize {
+    solve_part1(INPUT)
+}
+
+pub(crate) fn part2() -> usize {
+    solve_part2(INPUT)
+}
+
+type Grid = Vec<Vec<char>>;
+
+#[allow(clippy::identity_op)] // I consider the + 0 helpful for readability
+fn solve_part1(input: &str) -> usize {
+    let grid = parse(input);
+
+    let mut horizontal_count = 0;
+    for line in grid.iter() {
+        for w in line.windows(4) {
+            if w == ['X', 'M', 'A', 'S'] || w == ['S', 'A', 'M', 'X'] {
+                horizontal_count += 1;
+            }
+        }
+    }
+
+    let contains_xmas = |c1: &char, c2: &char, c3: &char, c4: &char| -> bool {
+        c1 == &'X' && c2 == &'M' && c3 == &'A' && c4 == &'S'
+            || c1 == &'S' && c2 == &'A' && c3 == &'M' && c4 == &'X'
+    };
+    let mut vertical_count = 0;
+    for x in 0..grid[0].len() {
+        for y in 0..grid.len() - 3 {
+            if contains_xmas(
+                &grid[y + 0][x],
+                &grid[y + 1][x],
+                &grid[y + 2][x],
+                &grid[y + 3][x],
+            ) {
+                vertical_count += 1;
+            }
+        }
+    }
+
+    let mut diagonal1_count = 0;
+    let mut diagonal2_count = 0;
+    for x in 0..grid[0].len() - 3 {
+        for y in 0..grid.len() - 3 {
+            if contains_xmas(
+                &grid[y + 0][x + 0],
+                &grid[y + 1][x + 1],
+                &grid[y + 2][x + 2],
+                &grid[y + 3][x + 3],
+            ) {
+                diagonal1_count += 1;
+            }
+            if contains_xmas(
+                &grid[y + 3][x + 0],
+                &grid[y + 2][x + 1],
+                &grid[y + 1][x + 2],
+                &grid[y + 0][x + 3],
+            ) {
+                diagonal2_count += 1;
+            }
+        }
+    }
+    horizontal_count + vertical_count + diagonal1_count + diagonal2_count
+}
+
+#[allow(clippy::identity_op)] // I consider the + 0 helpful for readability
+fn solve_part2(input: &str) -> usize {
+    let grid = parse(input);
+    let mut count = 0;
+    let contains_mas = |c1: &char, c2: &char, c3: &char| -> bool {
+        c1 == &'M' && c2 == &'A' && c3 == &'S' || c1 == &'S' && c2 == &'A' && c3 == &'M'
+    };
+    for x in 0..grid[0].len() - 2 {
+        for y in 0..grid.len() - 2 {
+            if contains_mas(
+                &grid[y + 0][x + 0],
+                &grid[y + 1][x + 1],
+                &grid[y + 2][x + 2],
+            ) && contains_mas(
+                &grid[y + 2][x + 0],
+                &grid[y + 1][x + 1],
+                &grid[y + 0][x + 2],
+            ) {
+                count += 1;
+            }
+        }
+    }
+    count
+}
+
+fn parse(input: &str) -> Grid {
+    input
+        .trim()
+        .lines()
+        .map(|line| line.chars().collect())
+        .collect()
+}
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    const EXAMPLE: &str = "\
+MMMSXXMASM
+MSAMXMSMSA
+AMXSXMAAMM
+MSAMASMSMX
+XMASAMXAMM
+XXAMMXXAMA
+SMSMSASXSS
+SAXAMASAAA
+MAMMMXMMMM
+MXMXAXMASX
+";
+
+    #[test]
+    fn test_part1_example() {
+        assert_eq!(18, solve_part1(EXAMPLE));
+    }
+
+    #[test]
+    fn test_part1() {
+        assert_eq!(2507, solve_part1(INPUT));
+    }
+
+    #[test]
+    fn test_part2_example() {
+        assert_eq!(9, solve_part2(EXAMPLE));
+    }
+
+    #[test]
+    fn test_part2() {
+        // not 1983, that's when also including a straight cross shape 🤦‍♂️
+        assert_eq!(1969, solve_part2(INPUT));
+    }
+}

2024/src/main.rs

@@ -3,11 +3,13 @@ use std::fmt::Display;
 mod day01;
 mod day02;
 mod day03;
+mod day04;
 
 fn main() {
     print_result(1, day01::part1(), day01::part2());
     print_result(1, day02::part1(), day02::part2());
     print_result(1, day03::part1(), day03::part2());
+    print_result(1, day04::part1(), day04::part2());
 }
 
 fn print_result(day: i32, part1: impl Display, part2: impl Display) {