Finding the first stream of non-repeating elements in Scala (without recursion or side-effects)

Here are some examples:

[1, 2, 3, 4, 5] => [1, 2, 3, 4, 5]
[10, 15, 10, 15, 30] => [10, 15]
[1, 2, 3, 4, 1, 5, 6, 7] => [1, 2, 3, 4]

Here's my best (and deeply ugly) non-recursive, side-effect-free solution so far:

x.scanLeft(List[Int]())((B, Term) => Term :: B).drop(1).takeWhile(i => !(i.tail contains i.head)).last.reverse

Minor optimization:

x.tail.scanLeft(List(x.head))((B, Term) => Term :: B).takeWhile(i => !(i.tail contains i.head)).last.reverse

This is different from distinct:

[1, 2, 3, 4, 1, 5, 6, 7] => [1, 2, 3, 4] and not [1, 2, 3, 4, 5, 6, 7]

Also, considering List[_] is a monoid, isn't there a scan that uses the monoid zero?

This is a fold, not a scan. A scan produces something with the same number of elements, and change the elements. A fold produces something new.

def firstDistinct[T](s: Seq[T]) = s.foldLeft(Seq[T]() -> false) {
  case (result @ (_, true), _)           => result
  case ((seq, _), el) if seq contains el => seq -> true
  case ((seq, _), el)                    => (seq :+ el) -> false
}._1

StackExchange Code Review Q#7200, answer score: 5