Get list of elements that are divisible by 3 or 5 from 1 - 1000

Question

I\'m trying to write a functional approach in scala to get a list of all numbers between 1 & 1000 that are divisible by 3 or 5

Here is what I have so far :

Answer 1

Looks like Brian beat me to it :)

Just thought I'd mention that a Stream might be more preferable here for better performance:

val x = (1 until 1000).toStream           //> x  : scala.collection.immutable.Stream[Int] = Stream(1, ?)
x filter (t=>(t%3==0)||(t%5==0))          //> res0: scala.collection.immutable.Stream[Int] = Stream(3, ?)

Answer 2

Here's how I would do it with a for expression.

for( i <- 1 to 1000 if i % 3 == 0 || i % 5 == 0) yield i

This gives:

 scala.collection.immutable.IndexedSeq[Int] = Vector(3, 5, 6, 9, 10, 12, 15, 18, 20, 21...

Here's another approach filtering on a Range of numbers.

scala> 1 to 1000
res0: scala.collection.immutable.Range.Inclusive = Range(1, 2, 3, 4, 5, 6, 7, 8, 9, 10...


scala> res0.filter(x => x % 3 == 0 || x % 5 == 0)
res1: scala.collection.immutable.IndexedSeq[Int] = Vector(3, 5, 6, 9, 10, 12, 15, 18, 20, 21...

If you really want a List on the return value use toList. e.g. res0.toList.

Answer 3

(Range(3, 1000, 3) ++ Range(5, 1000, 5)).toSet.toList.sorted

Sorted can be omitted.

Answer 4

The problem from projecteuler.net also wants a sum of those numbers at the end.

"Find the sum of all the multiples of 3 or 5 below 1000."

object prb1 {
  def main(args: Array[String]) {
    val retval = for{ a <- 1 to 999
                      if a % 3 == 0 || a % 5 == 0
    } yield a
    val sum = retval.reduceLeft[Int](_+_)
    println("The sum of all multiples of 3 and 5 below 1000 is " + sum)
  }
}

The correct answer should be 233168

Answer 5

No any answer without division or list recreation. No any answer with recursion.

Also, any benchmarking?

@scala.annotation.tailrec def div3or5(list: Range, result: List[Int]): List[Int] = {
  var acc = result
  var tailList = list
  try {
    acc = list.drop(2).head :: acc   // drop 1 2 save 3
    acc = list.drop(4).head :: acc   // drop 3 4 save 5
    acc = list.drop(5).head :: acc   // drop 5 save 6 
    acc = list.drop(8).head :: acc   // drop 6 7 8 save 9
    acc = list.drop(9).head :: acc   // drop 9 save 10
    acc = list.drop(11).head :: acc  // drop 10 11 save 12
    acc = list.drop(14).head :: acc  // drop 12 13 14 save 15 
    tailList = list.drop(15)         // drop 15             
  } catch {
    case e: NoSuchElementException => return acc // found
  }
  div3or5(tailList, acc) // continue search  
}

div3or5(Range(1, 1001), Nil)

EDIT

scala> val t0 = System.nanoTime; div3or5(Range(1, 10000001), Nil).toList; 
(System.nanoTime - t0) / 1000000000.0
t0: Long = 1355346955285989000
res20: Double = 6.218004

One of answers that looks good to me:

scala> val t0 = System.nanoTime; Range(1, 10000001).filter(i => 
i % 3 == 0 || i % 5 == 0).toList; (System.nanoTime - t0) / 1000000000.0
java.lang.OutOfMemoryError: Java heap space

Another one:

scala> val t0 = System.nanoTime; (Range(1, 10000001).toStream filter (
(t: Int)=>(t%3==0)||(t%5==0))).toList ; (System.nanoTime - t0) / 1000000000.0
java.lang.OutOfMemoryError: Java heap space

First one:

scala> val t0 = System.nanoTime; (for( i <- 1 to 10000000 if i % 3 == 0 || 
i % 5 == 0) yield i).toList; (System.nanoTime - t0) / 1000000000.0
java.lang.OutOfMemoryError: Java heap space

Why Scala does not optimize for example Vector -> List?

Answer 6

another aproach:

(1 to 1000).filter(i => i % 3 == 0 || i % 5 == 0)