Three Sum to N in Scala - scala

Question

Is there a better way than this example to find three numbers from a list that sum to zero in scala? Right now, I feel like my functional way may not be the most efficient and it contains duplicate tuples. What is the most efficient way to get rid of duplicate tuples in my current example?
def secondThreeSum(nums:List[Int], n:Int):List[(Int,Int,Int)] = {
val sums = nums.combinations(2).map(combo => combo(0) + combo(1) -> (combo(0), combo(1))).toList.toMap
nums.flatMap { num =>
val tmp = n - num
if(sums.contains(tmp) && sums(tmp)._1 != num && sums(tmp)._2 != num) Some((num, sums(tmp)._1, sums(tmp)._2)) else None
}
}

Answer 1

This is pretty simple, and doesn't repeat any tuples:
def f(nums: List[Int], n: Int): List[(Int, Int, Int)] = {
for {
(a, i) <- nums.zipWithIndex;
(b, j) <- nums.zipWithIndex.drop(i + 1)
c <- nums.drop(j + 1)
if n == a + b + c
} yield (a, b, c)
}

Answer 2

Use .combinations(3) to generate all distinct possible triplets of your start list, then keep only those that sum up to n :
scala> def secondThreeSum(nums:List[Int], n:Int):List[(Int,Int,Int)] = {
nums.combinations(3)
.collect { case List(a,b,c) if (a+b+c) == n => (a,b,c) }
.toList
}
secondThreeSum: (nums: List[Int], n: Int)List[(Int, Int, Int)]
scala> secondThreeSum(List(1,2,3,-5,2), 0)
res3: List[(Int, Int, Int)] = List((2,3,-5))
scala> secondThreeSum(List(1,2,3,-5,2), -1)
res4: List[(Int, Int, Int)] = List((1,3,-5), (2,2,-5))

Answer 3

Here is a solution that's O(n^2*log(n)). So it's quite a lot faster for large lists.
Also it uses lower level language features to increase the speed even further.
def f(nums: List[Int], n: Int): List[(Int, Int, Int)] = {
val result = scala.collection.mutable.ArrayBuffer.empty[(Int, Int, Int)]
val array = nums.toArray
val mapValueToMaxIndex = scala.collection.mutable.Map.empty[Int, Int]
nums.zipWithIndex.foreach {
case (n, i) => mapValueToMaxIndex += (n -> math.max(i, (mapValueToMaxIndex.getOrElse(n, i))))
}
val size = array.size
var i = 0
while(i < size) {
val a = array(i)
var j = i+1
while(j < size) {
val b = array(j)
val c = n - b - a
mapValueToMaxIndex.get(c).foreach { maxIndex =>
if(maxIndex > j) result += ((a, b, c))
}
j += 1
}
i += 1
}
result.toList
}