Day 10 part 2

solutions/day10.go

@@ -2,6 +2,7 @@ package solutions
 
 import (
 	"fmt"
+	"math/big"
 	"os"
 	"slices"
 )
@@ -24,8 +25,6 @@ func Day10Part01(isTest bool) int {
 		switch difference {
 		case 1:
 			jolt1Difference += 1
-		case 2:
-			continue
 		case 3:
 			jolt3Difference += 1
 		default:
@@ -38,6 +37,116 @@ func Day10Part01(isTest bool) int {
 	return jolt1Difference * jolt3Difference
 }
 
+// This section assumes that there are no 2-jolt difference connections.
+// In my input and the test inputs there were no 2-jolt differences.
 func Day10Part02(isTest bool) int {
+	numbers := ReadAndMapInt("10", isTest)
+	slices.Sort(numbers)
+
+	// The highest jolt value
+	highestValue := numbers[len(numbers)-1]
+	// Add the built-in adapter jolt value
+	numbers = append(numbers, highestValue+3)
+
+	// 1. Group all groups of 3 or more consecutive numbers.
+	counts := groupConsecutiveNumbers(numbers)
+
+	// 2. Count the number of paths between these groups
+	pathCounts := countPathsByConsecutiveNumbers(counts)
+
+	// 3. Multiply these number of paths to get the final answer.
+	answer := big.NewInt(1)
+	for _, value := range pathCounts {
+		answer.Mul(answer, big.NewInt(value))
+	}
+
+	fmt.Println("day 10 answer: " + answer.Text(10))
+
 	return -1
 }
+
+// Returns the length of each group of 3 or more consecutive numbers on the input slice.
+//
+// E.g.: This input:
+//
+//	0-3-4-5-9-12-13-14-15-16-19-20
+//
+// Would split into:
+//
+//	0 3-4-5 9 12-13-14-15-16 19 20
+//
+// And then return a slice:
+// [3, 5]
+// Which contains the length of each group of 3 or more consecutive numbers
+func groupConsecutiveNumbers(input []int) []int {
+	result := make([]int, 0)
+
+	previousNumber := 0
+	consecutiveCount := 1
+
+	for _, value := range input {
+		if value == previousNumber+1 {
+			consecutiveCount += 1
+		} else {
+
+			// If there was a consecutive group previously, append to the slice
+			if consecutiveCount >= 3 {
+				result = append(result, consecutiveCount)
+			}
+
+			// Reset the consecutive count
+			consecutiveCount = 1
+		}
+
+		previousNumber = value
+	}
+
+	return result
+}
+
+// Transforms a slice of amount of consecutive numbers into a slice of path counts
+//
+// Let x be a single amount of consecutive numbers:
+//
+// The amount of paths in a group of x numbers equals
+// the amount of paths in groups x-1 + x-2 + x-3.
+//
+// The initial state is as follows:
+//
+// - when x=1 then paths=1
+//
+// - when x=2 then paths=1
+//
+// - when x=3 then paths=2
+//
+// - when x=2 then paths = 2+1+1 = 4, and so on.
+//
+// This operation is mapped over every number in the input slice
+func countPathsByConsecutiveNumbers(input []int) []int64 {
+	// Initial state for memoization
+	pathCount := []int64{1, 1, 2, 4, 7, 13, 24}
+	result := make([]int64, len(input))
+
+	for idx, value := range input {
+		result[idx] = consecutiveCountToPathCount(value, &pathCount)
+	}
+
+	return result
+}
+
+func consecutiveCountToPathCount(consecutiveCount int, pathCountSlice *[]int64) int64 {
+	currentMaxAmount := len(*pathCountSlice)
+
+	// If the index is in the slice, return it
+	if currentMaxAmount >= consecutiveCount {
+		return (*pathCountSlice)[consecutiveCount-1]
+	}
+
+	// Otherwise, build the slice up to the index
+	for i := currentMaxAmount; i < consecutiveCount; i += 1 {
+		nextValue := (*pathCountSlice)[consecutiveCount-1] + (*pathCountSlice)[consecutiveCount-2] + (*pathCountSlice)[consecutiveCount-3]
+		*pathCountSlice = append(*pathCountSlice, nextValue)
+	}
+
+	return (*pathCountSlice)[consecutiveCount-1]
+}