Advent of Code 2023: Day 1

It’s that time of year again!

This year, I’ll be attempting the Advent of Code in Kotlin. This is a new language for me: Python is my default, and I attempted 2021 in Golang before ultimately switching back to Python.

I chose Kotlin because of its beautiful syntax for higher-order functions (map, forEach, filter, etc.). I find that most AoC problems boil down to processing a huge list, so those functions might be quite helpful. I certainly missed them when I used Go.

Code Structure

Since this is Day 1, I’ll walk you through the repo I’ve built. You can check out the code here.


All the solution code lives in src/main/kotlin. Here, I’ve built a base class, in BasePuzzle.kt. This class provides:

Each day’s puzzle gets its own class file, which extends BasePuzzle. These child classes are responsible for defining the input and output types, and implementing the methods for processing input and calculating the puzzle solution.

Each child class also has a companion object, with a main function, so you can run the code in IntelliJ IDEA with just one click.

companion object {
    @JvmStatic fun main(args: Array<String>) {
        val dayClass =
        val day = dayClass.newInstance() as BasePuzzle<*, *, *>

This main function remains the same for every puzzle, so it can be copy/pasted every day without modification. It simply finds the class that the companion object is attached to, creates a new instance, and calls the class’s main function.


Tests are defined in JUnit 5, in src/test/kotlin. There’s not much interesting to say about them, other than that the the test directory has its own resources directory. This allows us to use different input when testing instead of running the solution.

Now to get to the actual solution for Day 1. Spoilers ahead!

Day 1 code:

Reading the input

The input here is pretty straightforward. We’re interested in each individual line from the input file, which readLines already gives us. We just need to filter out blank lines, in case there’s one at the end of the file or something.

override fun getPuzzleInput(): List<String> {
    return readLines("day01.txt").filter { it.isNotBlank() }

In this case, our Input Type is a list of strings.

Part 1

Part 1 asks us to find the first and last digits (1 through 9) in each line. These can be the same digit, as in the fourth example: treb7uchet only contains a single 7, so that is both our first and last digit. Then we need to combine those digits in to a two-digit number for each line, and sum them all up.

override fun part1(input: List<String>): Int {
    return { line ->
        line.filter { it.isDigit() }
    }.map { line ->
        line.first().toString() + line.last().toString()
    }.sumOf { line ->

Kotlin’s iteration features are already coming in handy here. First off, we map over each line, filtering its characters down to only the digits. For the example input:

Next, we map again to get the start and end of each line, and concatenate them together into a string:

Finally, we use sumOf to convert these strings into integers, and add them all together to get our final result. sumOf combines both map and sum into one convenient method.

Part 2

Honestly, this is the trickiest Day 1 Part 2 puzzle I think I’ve ever seen. Some of the digits are spelled out, instead of written as digits. We now have to consider them, as well.

For my first attempt, I tried a rather messy find-and-replace to turn words into digits. Then, I reused the Part 1 code to do the actual calculation:

override fun part2(input: List<String>): Int {
    val parsedInput = {
        it.replace("one", "1")
            .replace("two", "2")
            .replace("three", "3")
            .replace("four", "4")
            .replace("five", "5")
            .replace("six", "6")
            .replace("seven", "7")
            .replace("eight", "8")
            .replace("nine", "9")
    return part1(parsedInput)

For the new test input:

Wait, that’s not right. Look at how we handle eightwo in the second input, and twone in the fourth. In both these cases, two digits share a letter, so only one can be converted. The find-and-replace is giving lower digits precedence over higher ones, so we favor the two in one case and the one in the other. But is that correct? No, the puzzle input is looking for the first and last digit in each line, regardless of its value.

Back to the drawing board

For our second attempt, we need to read the string in the proper order, stopping when we hit either a digit, or the name of a digit.

For the first digit, we want to check if the beginning of the line is a valid digit, and return it if so. Otherwise, we want to chop characters off of the beginning of the string, one at a time, until we find a digit.

We can boil the “check” down into a function, findDigitName.

private fun findDigitName(input: String): Int {
    if (input.isBlank())
        return 0

    if (input.startsWith("one"))
        return 1
    if (input.startsWith("two"))
        return 2
    if (input.startsWith("three"))
        return 3
    if (input.startsWith("four"))
        return 4
    if (input.startsWith("five"))
        return 5
    if (input.startsWith("six"))
        return 6
    if (input.startsWith("seven"))
        return 7
    if (input.startsWith("eight"))
        return 8
    if (input.startsWith("nine"))
        return 9

    if (input.first().isDigit())
        return input.first().digitToInt()

    return 0

For checking the first digit, we can iterate like so:

for (i in it.indices) {
    firstDigit = findDigitName(it.substring(i))
    if (firstDigit > 0) {

This looks at each index in the string, and evaluates it as if no characters before that index existed. Once it finds a substring that starts with a digit, it grabs that digit and stops iterating. For the third sample input, abcone2threexyz, it looks like this:

For finding the last digit, we just have to read backwards:

for (i in it.indices.reversed()) {
    lastDigit = findDigitName(it.substring(i))
    if (lastDigit > 0) {

Now that we have the first and last digits, we just have to turn them into a number. I realized at this point that we can do this mathematically, instead of with string concatenation.

firstDigit * 10 + lastDigit

All of this logic is wrapped in a sumOf, which gets us the correct final result!

See you tomorrow for Day 2!