Advent of Code 2023 - Day 1

It’s December and a new Advent of Code is ready to be solved.

This year I’m picking Rust, again. Let’s go ahead and solve Day 1!

You can find the final code @ advent-of-code/2023/day1

Our input is a multi-line text file, each line has variable length and consists of a sequence of chars and digits, here’s the example that is given to us


The problem goes on by saying that each line contains a calibration value that we need to get. In order to obtain that value, we have to combine the first digit and the last digit that we find on each line.

Let’s take the second line for example: if we walk every char we encounter 3 as the first digit and 8 as last digit, therefore, the calibration value is 38.

What about the last line? Well, we only have a single digit, which is 7, and that is both our first and last digit. The calibration value for the last line is 77.

This problem is pretty trivial, just what we wanted to warm ourselves for the 24 days ahead.

We have to start by reading the entire input file. It could be a useful exercise to code an input reader directly in the program but I like to handle this part by just piping the input into stdin with cat.

In Rust we can just create a mutable String and inject the content of stdin into it.

use std::io::{self, Read};

fn main() -> io::Result<()> {
    let mut input = String::new();
    io::stdin().read_to_string(&mut input);
    println!("{}", input);

We need to import std::io::Read to use read_to_string, otherwise the compiler will complain that that method is not implemented by stdin.

You can now run this command to see that everything is working as expected:

$ cat input | cargo run -

Now that we have our input, I want to create a function that takes an input and that is returns all the logic for part 1.

fn main() -> io::Result<()> {
    let mut input = String::new();
    let _ = io::stdin().read_to_string(&mut input);

    writeln!(io::stdout(), "{}", part1(&input)?)?;

fn part1(input: &str) -> io::Result<u32> {

Everything is setup, now we need to work on the solution.

What I would do in this case is:

  1. Iterate over each line

  2. Map each line to a the corresponding calibration value

  3. Sum all the values

fn part1(input: &str) -> io::Result<()> {
    let sum: u32 = input
        .map(|line| {
            // From the line we get an array of `char` and we try to parse each
            // one into a digit with `to_digit`.
            let digits: Vec<u32> = line
                // `to_digit` returns an `Option<u32>` and sice we only need valid
                // digits we can get rid of all the `None` values by using
                // `flat_map`.
                .flat_map(|x| x.to_digit(10))

            // We parse the calibration value by taking the first and last digit
            (digits[0] * 10) + digits[digits.len() - 1]

    writeln!(io::stdout(), "{}", sum)?;

To make sure that we did everything correctly I usually do a quick and dirty test

mod tests {
    use super::*;

    fn test_part1() {
        assert_eq!(part1("ad1f3").unwrap(), 13);
        assert_eq!(part1("ad1\na3\n11\n0").unwrap(), 11 + 33 + 11 + 0);

By running cargo test everything seems to pass, and if we paste the output solution returned by cat input | cargo run -, the AoC portal tells us that the solution is indeed correct.

We now have access to the second part of the problem: we are told that some of the digits are actually spelled with letters. For example, 7pqrstsixteen has a calibration value of 76 because the first digit is 7 and the last one is six.

We could reuse our part1 function if we could somehow replace spelled digits with its digit char, but a simple replace function is not going to do it in this case. Take a look at eightwothree for example, you can have different outcomes depending on the order of replacement that you apply:

  1. You may end up with 8wo3 if you replace eight before two

  2. You may end up with eigh23 if you replace two before eight

The problem description tells us that in this case the correct solution would be 8wo3, so I guess that we need to replace spelled digits in order of occurrence.

In this case, we can’t easily reuse the part1 function, but don’t worry, we can create a brand new function to solve this second part of the problem. I would solve this second part by doing the following:

  1. For each line find the indices of each digit or spelled digit

  2. Keep track of max digit and min digit

  3. Map line with found digit at min and max position

  4. Sum all the values

Rust has a nice function that returns all the indices of a certain substring in a string: match_indices, we can make use of that.

Here’s a quick sketch of the solution

fn repl_digits(x: &str) -> u32 {
    let digits = vec![
        ("1", "1"),
        ("2", "2"),
        ("3", "3"),
        ("4", "4"),
        ("5", "5"),
        ("6", "6"),
        ("7", "7"),
        ("8", "8"),
        ("9", "9"),
        ("0", "0"),
        ("one", "1"),
        ("two", "2"),
        ("three", "3"),
        ("four", "4"),
        ("five", "5"),
        ("six", "6"),
        ("seven", "7"),
        ("eight", "8"),
        ("nine", "9"),
        ("zero", "0"),

    // Initialize the first and last occurrences
    let mut first: (&str, usize) = ("", x.len());
    let mut last: (&str, usize) = ("", 0);

    for (substr, digit) in &digits {
        // Get the array of indices for the current substring
        let occ: Vec<_> = x.match_indices(substr).map(|x| x.0).collect();
        // If there's no occurrence, move to the next substr
        if occ.len() == 0 {

        // The indices array is ordered and we only need the first and last
        // occurrences
        let (min, max) = (occ[0], occ[occ.len() - 1]);

        // If the min index found is lte the current min index
        // keep track of the new first digit and its index
        if min <= first.1 {
            first = (digit, min);

        // If the max index found is gte the current max index
        // keep track of the new last digit and its index
        if max >= last.1 {
            last = (digit, max);

    // Parse value resulting by concatenating the two
    // digits
    format!("{}{}", first.0, last.0)

fn part2(input: &str) -> io::Result<u32> {
    let sum: u32 = input.lines().map(repl_digits).sum();


Quick test to make sure that everything is working properly

mod tests {
    use super::*;

    fn test_repl_digits() {
        assert_eq!(repl_digits("eightwothree"), 83);
        assert_eq!(repl_digits("13eightwothree"), 13);
        assert_eq!(repl_digits("13oneight"), 18);

    fn test_part2() {
        assert_eq!(part2("ad1f3").unwrap(), 13);
        assert_eq!(part2("ad1\na3\n11\n0").unwrap(), 11 + 33 + 11 + 0);
        assert_eq!(part2("zero").unwrap(), 0);
        assert_eq!(part2("three").unwrap(), 33);
        assert_eq!(part2("1\nthree\nonetwothree2three").unwrap(), 11 + 33 + 13);

cargo test gives us a green light, and indeed the solution is correct according to the AoC portal.

Yay, day 1 is done!