Scala - Combining two sequences to consecutively increasing triples - scala

Question

What is a nice and efficient functional way of solving the following problem? In imperative style, this can be done in linear time.
Given two sorted sequences p and q, f returns a sequence r (or any collection) of triples where for every triple (a,b,c) in r, the following hold:
(a < b < c)
One of the following two holds:
a,c are two consecutive elements p, and b is in q
a,c are two consecutive elements q, and b is in p
Example: Consider the following two sequences.
val p = Seq(1,4,5,7,8,9)
val q = Seq(2,3,6,7,8,10)
Then f(p,s) computes the following sequence:
Seq((1,2,4), (1,3,4), (5,6,7), (3,4,6), (3,5,6), (8,9,10))
Current solution: I do not find this one very elegant. I am looking for a better one.
def consecutiveTriplesOneWay(s1: Seq[Int], s2:Seq[Int]) = {
for {
i <- 0 until s1.size - 1 if s1(i) < s1(i+1)
j <- 0 until s2.size if s1(i) < s2(j) && s2(j) < s1(i+1)
} yield (s1(i), s2(j), s1(i+1))
}
def consecutiveTriples(s1: Seq[Int], s2:Seq[Int]) =
consecutiveTriplesOneWay(s1, s2) ++ consecutiveTriplesOneWay(s2, s1)
def main(args: Array[String]) {
val p = Seq(1,4,5,7,8,9)
val q = Seq(2,3,6,7,8,10)
consecutiveTriples(p, q).foreach(println(_))
}
Edit: My imperative solution
def consecutiveTriplesOneWayImperative(s1: Seq[Int], s2:Seq[Int]) = {
var i = 0
var j = 0
val triples = mutable.MutableList.empty[(Int,Int,Int)]
while (i < s1.size - 1 && j < s2.size) {
if (s1(i) < s2(j) && s2(j) < s1(i + 1)) {
triples += ((s1(i), s2(j), s1(i + 1)))
j += 1
} else if (s1(i) >= s2(j))
j += 1
else
i += 1
}
triples.toSeq
}
def consecutiveTriples(s1: Seq[Int], s2:Seq[Int]) =
consecutiveTriplesOneWayImperative(s1,s2) ++
consecutiveTriplesOneWayImperative(s2,s1)

Answer 1

Imperative solution translated to tailrec. Bit verbose but works
def consecutiveTriplesRec(s1: Seq[Int], s2: Seq[Int]) = {
#tailrec
def consTriplesOneWay(left: Seq[Int], right: Seq[Int],
triples: Seq[(Int, Int, Int)]): Seq[(Int, Int, Int)] = {
(left, right) match {
case (l1 :: l2 :: ls, r :: rs) =>
if (l1 < r && r < l2) consTriplesOneWay(left, rs, (l1, r, l2) +: triples)
else if (l1 >= r) consTriplesOneWay(left, rs, triples)
else consTriplesOneWay(l2 :: ls, right, triples)
case _ => triples
}
}
consTriplesOneWay(s1, s2, Nil) ++ consTriplesOneWay(s2, s1, Nil)
}