Check Is Anagram Scala HashMap based implementation

Question

Here is the HashMap based implementation of the function checking whether two strings are anagrams:

Example:

isAnagram("SAVE", "VASE") // returns true
isAnagram("SAV",  "VASE") // returns false
isAnagram("SAVE", "VAS")  // returns false
isAnagram("SAVE", "VAEE") // returns false

My implementation:

def isAnagram(value1: String, value2: String): Boolean = {
  value1.length == value2.length && {
    val index = new mutable.HashMap[Char, Int]()

    def addToIndex(c: Char) = index(c) = index.getOrElse(c, 0) + 1

    value1.foreach(c => addToIndex(c))

    def decrementAndCheckInIndex(c: Char): Boolean = index.get(c) match {
      case None => false
      case Some(count) => index(c) = count - 1; true
    }

    value2.forall(decrementAndCheckInIndex) && index.forall { case (key, value) => value == 0 }
  }
}

Could you please provide feedback on this code. I am interested in the mutable data structure use here, is it ok? And in general, what can I make better?

jwvh · Accepted Answer · 2018-03-06 09:26:24Z

It seems to me that groupBy() will do something very similar to your addToIndex(). The result is an immutable.Map[Char,String] instead of a mutable.HashMap[Char, Int], so the question is: which is better/easier for comparison purposes.

It turns out that the simple == evaluation is sufficiently deep to handle this.

def isAnagram(value1: String, value2: String): Boolean =
  value1.groupBy(identity) == value2.groupBy(identity)

Another, rather obvious, anagram test is simply:

value1.sorted == value2.sorted

I think, for long strings, grouping/indexing might be more efficient than sorting since each string is traversed only once. On the other hand, comparison for equality might be more complex for Maps than it would be for Strings.

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