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The problem is: Given two strings, write a method to decide if one is a permutation of the other. I wrote the following code in scala, may I know any other optimization one?

def checkpermutation(str1:String, str2:String): Boolean=(str1, str2) match {
   case (a,b) if a==b => true
   case (a,b) if a.length() !=b.length() =>false
   case (a,b) if a.toList.sorted.mkString== a.toList.sorted.mkString => true
   case _ =>false
 }
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  • \$\begingroup\$ The third case is invalid: it checks a.toList.sorted.mkString with itself. \$\endgroup\$ – Antot Nov 30 '16 at 12:43
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There is no need to use matchers for this check. A logical expression would be enough:

def checkpermutation2(a:String, b:String): Boolean = {
  def sorted(s: String) = s.sorted.mkString
  (a == b) || (a.length == b.length && sorted(a) == sorted(b))
} 
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