CODEWITHSUNDEEP
scalalistdata-structuresprimesfoldleft
hispaniccoder
hispaniccoder

Reputation: 397

How to return every nth even number in a list with foldLeft in Scala?

For a university project I have to implement a function called takeNthEven which finds the nth even number in a list with the aid of foldLeft. For example:

takeNthEven(SinglyLinkedIntList(0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15,18,5,3), 3)

should return: SinglyLinkedIntList(4, 10,18)

My attempt so far: 

def takeNthEven(input: IntList, n: Int): IntList = {
   var temp = SinglyLinkedIntList()

    input.foldLeft(0 -> 0) {
      case(acc,n) => if (acc == 2 || !(2 to (acc-1)).exists(x => i % x == 0)) temp.append(acc)
    }._n

  }

But unfortunately this does not even compile. I am not sure how to continue with this algorithm, could someone help me figure this out? I am new to functional programming and dont know how to go about this

Upvotes: 1

Views: 208

Answers (3)

stack0114106
stack0114106

Reputation: 8711

Another foldLeft

val n = 3
val p = List(0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15,18,5,3)

p.foldLeft( (List.empty[Int],1) ) {
  case((x,y),z) =>  {
    val t = if(y%n==0 && z%2==0 ) z::x else x
    (t,if(z%2==0) y+1 else y)
  }
}._1.reverse

// List[Int] = List(4, 10, 18)

Upvotes: 0

jwvh
jwvh

Reputation: 51271

Here's my take on it. You'll probably have to make adjustments depending on how your IntList is composed.

type IntList = List[Int]

def takeNthEven(input: IntList, n: Int): IntList = input.foldLeft(List.empty[Int]->1){
  case ((acc,cnt),elem) if elem%2 < 1 => if (cnt%n < 1) (elem::acc, cnt+1)
                                         else           (acc, cnt+1)
  case (skip,_) => skip
}._1.reverse

Upvotes: 1

Luis Miguel Mej&#237;a Su&#225;rez
Luis Miguel Mej&#237;a Su&#225;rez

Reputation: 22850

Something like this should work:

val (_, result) = input.foldLeft(0 -> List.empty[Int]) {
  case ((count, acc), elem) =>
    if ((elem % 2) == 0) {
      val newCount = count + 1
      if (newCount == n) {
        0 -> (elem :: acc)
      } else {
        newCount -> acc
      }
    } else {
      count -> acc
    }
}
result.reverse

Upvotes: 2

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