foldLeft and reverse

object SecretHandshake {
  val SIGNS = Vector("wink", "double blink", "close your eyes", "jump")
  val REVERSE_SIGNS = 16

  def commands(orders: Int): Seq[String] = {
    val signs =
      (0 until SIGNS.length)
        .foldLeft(Seq(): Seq[String])((output, index) =>
          if ((1 << index & orders) != 0) output :+ SIGNS(index)
          else output
        )

    if ((orders & REVERSE_SIGNS) == 0) signs else signs.reverse
  }
}

This approach starts by defining a Vector for holding the signs, and an Int binding for the reverse value.

The command() method starts by defining a Range from 0 up to but not including the length of the signs Vector. Each number in the Range is passed to the foldLeft() method for an index value.

The foldLeft() is initialized with an empty Seq[String] for the output value. The lambda in foldLeft uses the bitwise shift left operator to see if the input number has a binary 1 in the same position as 1 shifted left for the index value. If so, then the element for that index in the signs Vector is appended to the output Seq. If not, the output Seq is passed as is to the next iteration of foldLeft().

After foldLeft() is done, the signs binding is set to the output Seq.

The function uses a ternary expression to return either the Seq as is or the Seq in reverse(), depending if the input number has a bit set in the same position as binary 16 (10000).

Example

If the input number is 26, its value in binary is 11010. The 1 bits are set at four from the right, three from the right and one from the right.

• The element in the signs Vector at index 1 is "double blink", so that would get appended to the output Seq.
• The element in the signs Vector at index 3 is "jump", so that would get appended to the output Seq, after "double blink".
• Index 4 is beyond the length of the Vector, but 1 shifted left 4 places is 10000, which is the binary value for 16, so "double blink" and "jump" would be reversed and returned from commands() as ("jump", "double blink").