this post was submitted on 03 Dec 2024
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Advent Of Code

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๐Ÿท - 2024 DAY 3 SOLUTIONS -๐Ÿท (self.advent_of_code)
submitted 1 day ago* (last edited 1 day ago) by CameronDev to c/advent_of_code
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Day 3: Mull It Over

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[โ€“] [email protected] 3 points 1 day ago (2 children)

Go

Part 1, just find the regex groups, parse to int, and done.

Part 1

func part1() {
	file, _ := os.Open("input.txt")
	defer file.Close()
	scanner := bufio.NewScanner(file)

	re := regexp.MustCompile(`mul\(([0-9]{1,3}),([0-9]{1,3})\)`)
	product := 0

	for scanner.Scan() {
		line := scanner.Text()
		submatches := re.FindAllStringSubmatch(line, -1)

		for _, s := range submatches {
			a, _ := strconv.Atoi(s[1])
			b, _ := strconv.Atoi(s[2])
			product += (a * b)
		}
	}

	fmt.Println(product)
}

Part 2, not so simple. Ended up doing some weird hack with a map to check if the multiplication was enabled or not. Also instead of finding regex groups I had to find the indices, and then interpret what those mean... Not very readable code I'm afraid

Part2

func part2() {
	file, _ := os.Open("input.txt")
	defer file.Close()
	scanner := bufio.NewScanner(file)

	mulRE := regexp.MustCompile(`mul\(([0-9]{1,3}),([0-9]{1,3})\)`)
	doRE := regexp.MustCompile(`do\(\)`)
	dontRE := regexp.MustCompile(`don't\(\)`)
	product := 0
	enabled := true

	for scanner.Scan() {
		line := scanner.Text()
		doIndices := doRE.FindAllStringIndex(line, -1)
		dontIndices := dontRE.FindAllStringIndex(line, -1)
		mulSubIndices := mulRE.FindAllStringSubmatchIndex(line, -1)

		mapIndices := make(map[int]string)
		for _, do := range doIndices {
			mapIndices[do[0]] = "do"
		}
		for _, dont := range dontIndices {
			mapIndices[dont[0]] = "dont"
		}
		for _, mul := range mulSubIndices {
			mapIndices[mul[0]] = "mul"
		}

		nextMatch := 0

		for i := 0; i < len(line); i++ {
			val, ok := mapIndices[i]
			if ok && val == "do" {
				enabled = true
			} else if ok && val == "dont" {
				enabled = false
			} else if ok && val == "mul" {
				if enabled {
					match := mulSubIndices[nextMatch]
					a, _ := strconv.Atoi(string(line[match[2]:match[3]]))
					b, _ := strconv.Atoi(string(line[match[4]:match[5]]))
					product += (a * b)
				}
				nextMatch++
			}
		}
	}

	fmt.Println(product)
}

[โ€“] [email protected] 2 points 1 day ago (1 children)

I also used Go - my solution for part 1 was essentially identical to yours. I went a different route for part 2 that I think ended up being simpler though.

I just prepended do() and don't() to the original regex with a |, that way it captured all 3 in order and I just looped through all the matches once and toggled the isEnabled flag accordingly.

Always interesting to see how other people tackle the same problem!

Part 2 Code

func part2() {
	filePath := "input.txt"
	file, _ := os.Open(filePath)
	defer file.Close()

	pattern := regexp.MustCompile(`do\(\)|don't\(\)|mul\((\d{1,3}),(\d{1,3})\)`)
	productSum := 0
	isEnabled := true

	scanner := bufio.NewScanner(file)
	for scanner.Scan() {
		line := scanner.Text()
		matches := pattern.FindAllStringSubmatch(line, -1)

		for _, match := range matches {
			if match[0] == "do()" {
				isEnabled = true
			} else if match[0] == "don't()" {
				isEnabled = false
			} else if isEnabled && len(match) == 3 {
				n, _ := strconv.Atoi(match[1])
				m, _ := strconv.Atoi(match[2])
				productSum += n * m
			}
		}
	}

	fmt.Println("Total: ", productSum)
}

[โ€“] [email protected] 1 points 1 day ago

Honestly this is soo much better, I'm not proud of my code at all haha. Thanks for sharing, definitely adding that | to my bag of tricks