foldLeft

Phone Number
Phone Number in Scala 
object PhoneNumber {

  def clean(input: String): Option[String] = {
    input
      .foldLeft((List[Char](), 0))((tup, head) =>
        tup._2 match {
          case index if (List(0, 3).contains(index)) => {
            head.isDigit match {
              case true if (head > '1') => (head :: tup._1, index + 1)
              case true if (index == 0 && head == '1') =>
                tup
              case false
                  if (index == 0 && List('+', '(', ' ').contains(head)) =>
                tup
              case false
                  if (index == 3 && List(')', ' ', '.').contains(head)) =>
                tup
              case _ => (Nil, -100)
            }
          }
          case index => {
            head.isDigit match {
              case true => (head :: tup._1, index + 1)
              case false
                  if (index == 6 && List(' ', '-', '.').contains(head)) =>
                tup
              case false if (head == ' ') => tup
              case _                      => (Nil, -100)
            }
          }
        }
      ) match {
      case (output, index) =>
        if (index == 10) Some(output.reverse.mkString) else None
    }
  }
}

This approach starts by calling the foldLeft() method in the input String. It is initialized with an emptyList for building the output String and 0 for the index, both wrapped in a tuple. The tuple and each character in the input String are passed to the lambda, where the tuple is passed to the match. The pattern matching checks if the index is 0 or 3.

If so, and the head character is a digit greater than 1, then a new tuple is made from the the head prepended to the output List with the cons operator (::), and with the index added to by 1. If the head is some other legal character for its index, then the tuple is passed as is. If the head is not a legal character for its index, then a new tuple is made from a Nil output List and with the index given a value such that it will reasonably never be valid.

If the index is neither 0 or 3, then the head is checked for being a digit. If it is a digit, then a new tuple is made from the the head prepended to the output List with the cons operator (::), and with the index added to by 1. If the head is not a digit but is some other legal character, then the tuple is passed as is. If the head is not a legal character, then a new tuple is made from a Nil output List and with the index given a value such that it will reasonably never be valid.

The result of foldLeft() is passed to the match. If the index is 10, then the reverse and mkString methods are used on the output List to return a Some value from the recursive method. If the index is not 10, then None is returned from the recursive method.

9th Oct 2024 · Found it useful?

Other Approaches to Phone Number in Scala

Other ways our community solved this exercise
case (head :: tail, index) => {
  head.isDigit match {
    case true => cleanRecur(tail, head :: cleaned)
    case false if (index == 6 && List(' ', '-', '.').contains(head)) =>
      cleanRecur(tail, cleaned)
    case false if (head == ' ') => cleanRecur(tail, cleaned)
    case _                      => None
}}
Recursion

Use recursion to return the answer. 

object PhoneNumber {
  private val VALID_PHONE_NUMBER = raw"^1?([2-9]\d{2}[2-9]\d{6})$$".r

  def clean(input: String): Option[String] = {
    val cleaned = input.filter(_.isDigit)
    Option(cleaned) collect { case VALID_PHONE_NUMBER(cleaned) => cleaned }
  }
}
Regex with filter

Use Regex with filter to return the answer.