Find first repeating Char in String












0

















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.






strings recursion interview-questions functional-programming scala







share|improve this question




















  • 1




    Well there seems to be a problem with either the problem-statement or with the examples.
    – Heslacher
    22 hours ago
















0

















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.






strings recursion interview-questions functional-programming scala







share|improve this question




















  • 1




    Well there seems to be a problem with either the problem-statement or with the examples.
    – Heslacher
    22 hours ago














0












0








0










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.






strings recursion interview-questions functional-programming scala







share|improve this 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.






strings recursion interview-questions functional-programming scala




strings recursion interview-questions functional-programming scala






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share|improve this question













share|improve this question




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edited 6 hours ago









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128k15152414










asked yesterday









vikrantvikrant

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  • 1




    Well there seems to be a problem with either the problem-statement or with the examples.
    – Heslacher
    22 hours ago














  • 1




    Well there seems to be a problem with either the problem-statement or with the examples.
    – Heslacher
    22 hours ago








1




1




Well there seems to be a problem with either the problem-statement or with the examples.
– Heslacher
22 hours ago




Well there seems to be a problem with either the problem-statement or with the examples.
– Heslacher
22 hours ago










1 Answer
1






active

oldest

votes


















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 })





share|improve this answer























  • thanks for this wonderful answer.
    – vikrant
    2 hours ago






  • 1




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













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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









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 })





share|improve this answer























  • thanks for this wonderful answer.
    – vikrant
    2 hours ago






  • 1




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


















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 })





share|improve this answer























  • thanks for this wonderful answer.
    – vikrant
    2 hours ago






  • 1




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
















1












1








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 })





share|improve this answer














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 })






share|improve this answer














share|improve this answer



share|improve this answer








edited 6 hours ago

























answered 6 hours ago









200_success200_success

128k15152414




128k15152414












  • thanks for this wonderful answer.
    – vikrant
    2 hours ago






  • 1




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




















  • thanks for this wonderful answer.
    – vikrant
    2 hours ago






  • 1




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


















thanks for this wonderful answer.
– vikrant
2 hours ago




thanks for this wonderful answer.
– vikrant
2 hours ago




1




1




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






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




















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