Advent of Code Using Tidyverse

When I find the time, I like to try solving the Advent of Code puzzles. I prefer to take a “tidy,” data-like approach wherever possible (because that’s all I know). Let’s see how far that gets me this year.

Day 2 asks us to find numbers that contain patterns (like 99 or 1188511885) within ranges of numbers. This task is very friendly to a tidyverse approach. We need to do some data wrangling to get everything in the right format, then it’s just a matter of a simple regex to find what we need.

The separate functions from tidyr and a little regex made today’s solution clean and straightforward.

library(tidyverse)

input <- "11-22,95-115,998-1012,1188511880-1188511890,222220-222224,1698522-1698528,446443-446449,38593856-38593862,565653-565659,824824821-824824827,2121212118-2121212124"

data <- tibble(input = input) |>
  separate_longer_delim(input, ",") |>
  separate(input, into = c("start", "end"), sep = "-") |>
  rowwise() |>
  mutate(range = list(start:end)) |>
  unnest_longer(range)

data

# A tibble: 106 × 3
   start end   range
   <chr> <chr> <int>
 1 11    22       11
 2 11    22       12
 3 11    22       13
 4 11    22       14
 5 11    22       15
 6 11    22       16
 7 11    22       17
 8 11    22       18
 9 11    22       19
10 11    22       20
# ℹ 96 more rows

range here gives us all the values in all the ranges. Now it’s just a matter of finding the repetitions using regex, and adding them all together.

 data |>
  filter(str_detect(range, "^([0-9]+)\\1$")) |>
  summarise(sum(range))

# A tibble: 1 × 1
  `sum(range)`
         <int>
1   1227775554

For part 2, all I needed to add was a + to the regex to allow for more than one repetition of the pattern.

# Part 2

tibble(input = input) |>
  separate_longer_delim(input, ",") |>
  separate(input, into = c("start", "end"), sep = "-") |>
  rowwise() |>
  mutate(range = list(start:end)) |>
  unnest_longer(range) |>
  # adding a plus here for unlimited repeats
  filter(str_detect(range, "^([0-9]+)\\1+$")) |>
  summarise(sum(range))

# A tibble: 1 × 1
  `sum(range)`
         <dbl>
1   4174379265