day 25 article

Author: merlinorg

docs/2024/puzzles/day25.md

@@ -2,10 +2,84 @@ import Solver from "../../../../../website/src/components/Solver.js"
 
 # Day 25: Code Chronicle
 
+by [@merlinorg](https://github.com/merlinorg)
+
 ## Puzzle description
 
 https://adventofcode.com/2024/day/25
 
+## Solution summary
+
+1. Parse and partition the locks and keys.
+2. Find all lock and key combinations that could fit.
+3. Return the count of potential matches.
+
+## Parsing
+
+It's the last day of Advent of Code so we'll keep it simple. The input consists
+of a sequence of grids, separated by blank lines. Each grid is a sequence of lines
+consisting of the `#` and `.` characters. Locks have `#` characters in the first
+row,  where keys have `.` characters. To parse this we'll just use simple string
+splitting.
+
+First, we split the input into grids by matching on the blank lines. This gives us
+an array of strings, each grid represented by a single string. Then we partition
+this into two arrays; one the keys, and the other locks. For this, we use the
+`partition` method that takes a predicate; every grid that matches this predicate
+will be placed in the first array of the resulting tuple, the rest in the second.
+
+```scala 3
+val (locks, keys) = input.split("\n\n").partition(_.startsWith("#"))
+```
+
+In general, arrays are not the most ergonomic data structures to use, but for
+this final puzzle they are more than sufficient.
+
+## Matching
+
+To find all potential matches we will use a for comprehension to loop through
+all the locks, and then for each lock, through all the keys. For each pair of
+a lock and key, we want to determine whether there is any overlap that would
+prevent the key fitting the lock.
+We can perform this test by simply zipping the key and the lock strings; this
+gives us a collection of every corresponding character from each string. The
+key can fit the lock if there is no location containing a `#` character in
+both grids.
+
+```scala 3
+val matches = for
+  lock <- locks
+  key  <- keys
+  if lock.zip(key).forall: (lockChar, keyChar) =>
+    lockChar != '#' || keyChar != '#'
+yield lock -> key
+```
+
+This returns all of the matching lock and key combinations; the solution to
+the puzzle is the size of this array.
+
+## Final code
+
+```scala 3
+def part1(input: String): Int =
+  val (locks, keys) = input.split("\n\n").partition(_.startsWith("#"))
+
+  val matches = for
+    lock <- locks
+    key  <- keys
+    if lock.zip(key).forall: (lockChar, keyChar) =>
+      lockChar != '#' || keyChar != '#'
+  yield lock -> key
+
+  matches.length
+```
+
+### Run it in the browser
+
+#### Part 1
+
+<Solver puzzle="day25-part1" year="2024"/>
+
 ## Solutions from the community
 
 Share your solution to the Scala community by editing this page.

solver/src/main/scala/adventofcode/Solver.scala

@@ -15,6 +15,7 @@ object Solver:
       "day21-part2" -> day21.part2,
       "day22-part1" -> day22.part1,
       "day22-part2" -> day22.part2,
+      "day25-part1" -> day25.part1,
     )
 
   private val solutions2023: Map[String, String => Any] =