Scala convert IndexedSeq[AnyVal] to Array[Int]

Question

I'm trying to solve Codility's GenomicRangeQuery using Scala and to that end I wrote the following function:

def solution(s: String, p: Array[Int], q: Array[Int]): Array[Int] = {
  for (i <- p.indices) yield {
    val gen = s.substring(p(i), q(i) + 1)
    if (gen.contains('A')) 1
    else if (gen.contains('C')) 2
    else if (gen.contains('G')) 3
    else if (gen.contains('T')) 4
  }
}

I haven't done a lot of testing but it seems to solve the problem.

My issue is the for comprehension returns an scala.collection.immutable.IndexedSeq[AnyVal], while the function must return an Array[Int] and therefore it's throwing a type mismatch error.

Is there any way to make the for comprehension return an Array[Int] or transform the IndexedSeq[AnyVal] into an Array[Int]?

Answer 1

sheunis' answer above mostly covers it.

You can coerce an IndexedSeq into an Array with a call to toArray so the first bit's quite straightforward. For the second part, because there's a possible logical branch where you drop through all of your if... else... cases, it's possible for your yield to return both Int and Unit types, whose closest common ancestor is AnyVal.

Note that if you replaced your if... else... with pattern matching instead then you would explicitly get a compiler warning because you're not catching every possible case.

gen match { case _ if gen.contains("A") => 1 case _ if gen.contains("C") => 2 ... // Throws warning unless you include a `case _ =>` with no `if` clause }

Answer 2

def solution(s: String, p: Array[Int], q: Array[Int]): Array[Int] = {
  (for (i <- p.indices) yield {
    val gen = s.substring(p(i), q(i) + 1)
    if (gen.contains('A')) 1
    else if (gen.contains('C')) 2
    else if (gen.contains('G')) 3
    else 4
  }).toArray
}

The problem with the if statement is that there is no default value, which is why you get an IndexedSeq of Any instead of Int.