Given a variable length array of integers, partition them such that the even integers precede the odd integers in the array. Your must operate on the array in-place, with a constant amount of extra space. The answer should scale linearly in time with respect to the size of the array.

EvenOdd.scala

object Main {
  def sortOddEven(numbers:Array[Int]) = {
    println(numbers.mkString(","))
    var left = 0
    var right = numbers.length - 1
    while (left < right) {
      //Increment left index till we see an odd
      while (numbers(left) % 2 == 0 && left < right)
        left+=1

      //Decrement right index till we see an even
      while (numbers(right) % 2 == 1 && left < right)
        right-=1

      //swap the odd and even positions
      if (left < right) {
        val temp = numbers(left)
        numbers(left) = numbers(right)
        numbers(right) = temp
        left+=1
        right-=1
      }
    }
    numbers
  }

  def main(args: Array[String]): Unit = {
    println(sortOddEven(args(0).split(",").map(_.toInt)).mkString(","))
  }
}

@binshi What's the source of this and Compute100 problems? Google job interview? Best regards