Solution article for 2024 day 02 (#566)

* Solution article for 2024 day 02

* fix some typos

Author: spamegg1

docs/2024/puzzles/day02.md

@@ -6,6 +6,85 @@ import Solver from "../../../../../website/src/components/Solver.js"
 
 https://adventofcode.com/2024/day/2
 
+## Solution summary
+
+- First we parse each line of the input into a `Report` case class: `case class Report(levels: Seq[Long])`
+- In each `Report`, we construct the sequence of consecutive pairs.
+- For part 1, we check if the pairs are all increasing or all decreasing, and if the difference is within the given limits (such a report is "safe").
+- For part 2, we construct new `Report`s obtained by dropping one entry in the original report, and check if there exists a safe `Report` among these.
+- In both parts, we simply count the number of `Report`s that are considered safe.
+
+### Parsing
+
+Each line of input is a string of numbers separated by a single space.
+Therefore parsing looks like this:
+
+```scala
+case class Report(levels: Seq[Long])
+
+def parseLine(line: String): Report = Report(line.split(" ").map(_.toLong).toSeq)
+
+def parse(input: String): Seq[Report] = input
+  .linesIterator
+  .map(parseLine)
+  .toSeq
+```
+
+### Part 1: methods of the `Report` case class
+
+We need to check consecutive pairs of numbers in each report in 3 ways:
+
+- to see if they are all increasing,
+- to see if they are all decreasing,
+- to see if their differences are within given bounds.
+
+So let's construct them only once, save it as a `val`, then reuse this value 3 times.
+It's not the most efficient way (like traversing only once and keeping track of everything), 
+but it's very clean and simple:
+
+```scala
+case class Report(levels: Seq[Long]):
+  val pairs = levels.init.zip(levels.tail) // consecutive pairs
+  def allIncr: Boolean = pairs.forall(_ < _)
+  def allDecr: Boolean = pairs.forall(_ > _)
+  def within(lower: Long, upper: Long): Boolean = pairs.forall: pair =>
+    val diff = math.abs(pair._1 - pair._2)
+    lower <= diff && diff <= upper
+  def isSafe: Boolean = (allIncr || allDecr) && within(1L, 3L)
+```
+
+Part 1 solver simply counts safe reports, so it looks like this:
+
+```scala
+def part1(input: String): Int = parse(input).count(_.isSafe)
+```
+
+### Part 2
+
+Now we add new methods to `Report`. 
+We check if there exists a `Report` obtained by dropping one number, such that it's safe.
+We do this by iterating over the index of each `Report`.
+Then, a `Report` is safe, if it's safe as in Part 1, or one of the dampened reports is safe:
+
+```scala
+case class Report(levels: Seq[Long]):
+  // ... as before
+  def checkDampenedReports: Boolean = (0 until levels.size).exists: index =>
+    val newLevels = levels.take(index) ++ levels.drop(index + 1)
+    Report(newLevels).isSafe
+  def isDampenedSafe: Boolean = isSafe || checkDampenedReports
+```
+
+Again this is not the most efficient way (we are creating many new `Report` instances), 
+but our puzzle inputs are fairly short (there are at most 8 levels in each `Report`), 
+so it's a simple approach that reuses the `isSafe` method from Part 1.
+
+Part 2 solver now counts the dampened safe reports:
+
+```scala
+def part2(input: String): Int = parse(input).count(_.isDampenedSafe)
+```
+
 ## Solutions from the community
 
 - [Solution](https://github.com/rmarbeck/advent2024/tree/main/day2) by [Raphaël Marbeck](https://github.com/rmarbeck)