1

Solved! Solution at the bottom.

I'm porting some Java code to Scala for fun and I trapped into a pretty nifty way of bit-shifting in Java. The Java code below takes a String as input and tests if it consists of unique characters. 

public static boolean isUniqueChars(String str) {
if (str.length() > 256)return false; }
int checker = 0;
 for (int i = 0; i < str.length(); i++) {
    int val = str.charAt(i) - 'a';
  if ((checker & (1 << val)) > 0) return false;
 checker |= (1 << val);
 }
return true;

Full listing is here: https://github.com/marvin-hansen/ctci/blob/master/java/Chapter%201/Question1_1/Question.java

How the code exactly works is explained here: How does this Java code which determines whether a String contains all unique characters work?

Porting this directly to Scala doesn't really work so I'm looking for a more functional way to re-write the stuff above.

I have tried BigInt & BitSet 

def isUniqueChars2(str : String) : Boolean =
// Java, char's are Unicode so there are 32768 values
if (str.length() > 32768)  false
val checker = BigInt(1)
 for(i <- 0 to str.length){
   val value = str.charAt(i)
   if(checker.testBit(value)) false
   checker.setBit(value)
 }
true
}

This works, however, but without bit-shifting and without lowercase assumption. Performance is rather unknown ....

However, I would like to do a more functional style solution.

Thanks to user3189923 for the solution. 

 def isUniqueChars(str : String) = str.distinct == str

That's it. Thank you.

Community
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Marvin.Hansen
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2 Answers2

2
str.distinct == str

In general, method distinct preserves order of occurrence after removing duplicates. Consider

implicit class RichUnique(val str: String) extends AnyVal {
  def isUniqueChars() = str.distinct == str
}

and so

"abc".isUniqueChars
res: Boolean = true

"abcc".isUniqueChars
res: Boolean = false
elm
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1

How about:

str.toSet.size == str.size

?

Noel M
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