Concat sorted generic sequence - scala

Question

I need to concat two generic sequences, I tried to do it like this, but I understand that this is wrong. How should I do it in right way? (I need to get new Seq which will be ordered too)
object Main extends App{
val strings = Seq("f", "d", "a")
val numbers = Seq(1,5,4,2)
val strings2 = Seq("c", "b");
val strings3 = strings2.concat(strings)
println(strings3)
println(numbers)
}
class Seq[T] private(initialElems: T*) {
override def toString: String = initialElems.toString
val elems = initialElems
def concat(a:Seq[T]) = a.elems ++ this.elems
}
object Seq {
def apply[T: Ordering](initialElems: T*): Seq[T] = new Seq(initialElems.sorted:_*)
}

Answer 1

You could create a function which will go through both lists taking heads of both lists and comparing them, then appending appropriate head to the result list. Then it should take the next two heads and repeat until one list is over.
Here's tail-recursive example:
import scala.annotation.tailrec
def merge[A](a: List[A], b: List[A])(implicit ordering: Ordering[A]): List[A] = {
#tailrec
def go(a: List[A], b: List[A], acc: List[A] = Nil): List[A] = {
(a, b) match {
case (ax :: as, bx :: bs) => if(ordering.compare(ax, bx) < 0) go(as, bx :: bs, ax :: acc) else go(ax :: as, bs, bx :: acc)
case (Nil, bs) => acc.reverse ++ bs
case (as, Nil) => acc.reverse ++ as
case _ => acc.reverse
}
}
go(a, b)
}
val strings = List("a", "d", "f")
val strings2 = List("b", "c")
merge(strings, strings2) // List(a,b,c,d,e)
I used List instead of Seq. You should rather not use Seq, which is very general type, but utilize more specific collection types, which suit your task best, like Vector, List, ArraySeq etc.

Answer 2

You can't concat two sorted arrays using ++ keeping order. ++ just stick one sequence to the end of another.
You need to implement something like merge operation from merge sort algorithm and create new Seq from merged elems without sorting.
So, you need to do 3 things:
Implement merge:
def merge(a: Seq[T], b: Seq[T]): YourElemsType[T] = ???
Implement new method for creating Seq instance without sorting in object Seq:
def fromSorted(initialElems: T*): Seq[T] = new Seq(initialElems:_*)
After all, your concat can be implemented as composition merge and fromSorted:
def concat(a:Seq[T]): Seq[T] = Seq.fromSorted(merge(this, a))
Read more about merge sort wiki