Find first repeating Char in String

Question

Given a string, find the first repeating character in it.

Examples:

• firstUnique("Vikrant") → None
• firstUnique("VikrantVikrant") → Some(V)

Scala implementation:

object FirstUniqueChar extends App {
 def firstUnique(s: String): Option[Char] = {
 val countMap = (s groupBy (c=>c)) mapValues(_.length)
 def checkOccurence(s1: String ): Option[Char] = {
 if (countMap(s1.head) > 1) Some(s1.head)
 else if (s1.length == 1) None
 else checkOccurence(s1.tail)
 }
 checkOccurence(s)
 }
 println(firstUnique("abcdebC"))
 println(firstUnique("abcdef"))
}

I also have a followup question. What is the recommended way if I do not want to solve this problem with recursion? Instead of using the checkOccurence method I can traverse through the string and break when I find the first element with a count more than 1. But that will require a break, which is discouraged in Scala.

Answer 1

Your checkOccurrence(s) is just a clumsy way to write s.find(countMap(_)> 1).

You can significantly simplify the solution by taking advantage of .distinct.

def firstUnique(s: String): Option[Char] =
 s.zipAll(s.distinct, '\u0000', '\u0000')
 .collectFirst({ case ab if ab._1 != ab._2 => ab._1 })

• jwvh: Slightly more concise syntax for a very nice answer: collectFirst{case (a,b) if a != b => a}