Day 3: Mull It Over

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FAQ

  • proved_unglue
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    1 day ago

    Kotlin

    fun part1(input: String): Int {
        val pattern = "mul\\((\\d{1,3}),(\\d{1,3})\\)".toRegex()
        var sum = 0
        pattern.findAll(input).forEach { match ->
            val first = match.groups[1]?.value?.toInt()!!
            val second = match.groups[2]?.value?.toInt()!!
            sum += first * second
    
        }
        return sum
    }
    
    fun part2(input: String): Int {
        val pattern = "mul\\((\\d{1,3}),(\\d{1,3})\\)|don't\\(\\)|do\\(\\)".toRegex()
        var sum = 0
        var enabled = true
        pattern.findAll(input).forEach { match ->
            if (match.value == "do()") enabled = true
            else if (match.value == "don't()") enabled = false
            else if (enabled) {
                val first = match.groups[1]?.value?.toInt()!!
                val second = match.groups[2]?.value?.toInt()!!
                sum += first * second
            }
        }
        return sum
    }
    
    • Faresh@lemmy.ml
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      1 day ago

      You can avoid having to escape the backslashes in regexps by using multiline strings:

      val pattern = """mul\((\d{1,3}),(\d{1,3})\)""".toRegex()