day1 part2

Author: quan-nh

resources/2024/day01-requirement.md

@@ -22,3 +22,35 @@ Example calculation:
 - Total distance: 2+1+0+1+2+5 = 11
 
 The puzzle asks you to perform this calculation on the actual input lists to determine the total distance.
+
+--- Part Two ---
+
+The puzzle now focuses on calculating a "similarity score" between two lists of numbers. The process is as follows:
+
+For each number in the left list:
+
+- Count how many times that number appears in the right list
+- Multiply the number by its frequency in the right list
+- Add this product to a running total (similarity score)
+
+Key points:
+
+- Numbers that don't appear in the right list contribute 0 to the score
+- The score is calculated by multiplying each left list number by its frequency in the right list
+- The goal is to find the total similarity score
+
+Example calculation:
+
+- Left list: [3, 4, 2, 1, 3, 3]
+- Right list: [4, 3, 5, 3, 9, 3]
+- Breakdown:
+    - 3 appears 3 times in right list: 3 * 3 = 9
+    - 4 appears 1 time in right list: 4 * 1 = 4
+    - 2 appears 0 times in right list: 2 * 0 = 0
+    - 1 appears 0 times in right list: 1 * 0 = 0
+    - 3 appears 3 times in right list: 3 * 3 = 9
+    - 3 appears 3 times in right list: 3 * 3 = 9
+
+- Total similarity score: 9 + 4 + 0 + 0 + 9 + 9 = 31
+
+The puzzle asks you to calculate this similarity score for the actual input lists.

src/adventofcode2024/day01.odin

@@ -21,12 +21,44 @@ main :: proc() {
 		append(&y, strconv.atoi(s[1]))
 	}
 
-	slice.sort(x[:])
-	slice.sort(y[:])
+	part1_result := part1(x[:], y[:])
+	fmt.println(part1_result) // 1765812
+
+	part2_result := part2(x[:], y[:])
+	fmt.println(part2_result) // 20520794
+}
+
+part1 :: proc(x, y: []int) -> int {
+	slice.sort(x)
+	slice.sort(y)
 
 	result := 0
 	for i in 0 ..< len(x) {
 		result += abs(x[i] - y[i])
 	}
-	fmt.println(result) // 1765812
+
+	return result
+}
+
+part2 :: proc(x, y: []int) -> int {
+	m: map[int]int
+	defer delete(m)
+
+	result := 0
+	for i in x {
+		if freq, ok := m[i]; ok {
+			result += i * freq
+		} else {
+			freq := 0
+			for j in y {
+				if i == j {
+					freq += 1
+				}
+			}
+			result += i * freq
+			m[i] = freq
+		}
+	}
+
+	return result
 }