Kotlin
I like this one. The first part could be solved trivially with a naive approach. The second part is also not too complicated.
I started out with a simple merging algorithm that would check all ranges, except the one to merge, for overlaps.
With clever sorting, I can skip the search for mergable ranges and just iterate in reverse and build larger and larger ranges.
I haven't tried to merge adjacent ranges like 1..2 and 3..4 to 1..4. The solution is fast enough already when JIT/C2 compiled (200-250 ยตs).
Code on Github
Code
class Day05 : AOCSolution {
override val year = 2025
override val day = 5
override fun part1(inputFile: String): String {
val (freshIngredients, availableIngredients) = readResourceTwoParts(inputFile)
val freshRanges = buildRanges(freshIngredients)
val ingredientIds = availableIngredients.lines().filterNot(String::isEmpty).map(String::toLong)
val count = ingredientIds.count { id -> freshRanges.any { range -> id in range } }
return count.toString()
}
override fun part2(inputFile: String): String {
val (freshIngredients, _) = readResourceTwoParts(inputFile)
val freshRanges = buildRanges(freshIngredients)
return freshRanges.sumOf { range ->
range.last - range.first + 1
}.toString()
}
private companion object {
private fun buildRanges(ingredients: String): List<LongRange> {
val lines = ingredients.lines()
val ranges = MutableList(lines.size) { i ->
val line = lines[i]
val hyphen = line.indexOf('-')
val lower = line.take(hyphen).toLong()
val upper = line.substring(hyphen + 1).toLong()
LongRange(lower, upper)
}
// Sort the ranges
// The ones with the smallest ID and the least upper end
// get sorted to the beginning.
// This allows for easy merging, as overlapping ranges are always adjacent
ranges.sortWith(Comparator { r1, r2 ->
val first = r1.first.compareTo(r2.first)
if (first != 0) {
first
} else {
r1.last.compareTo(r2.last)
}
})
// Merge adjacent ranges backwards, modifying the list in-place
for (i in ranges.lastIndex downTo 1) {
val lowerRange = ranges[i - 1]
val upperRange = ranges[i]
// The two ranges do overlap, because the tail of the first range
// is in the second range.
if (upperRange.first <= lowerRange.last) {
ranges[i - 1] = LongRange(
lowerRange.first,
maxOf(lowerRange.last, upperRange.last)
)
ranges.removeAt(i)
}
}
return ranges
}
}
}