Lexicographically kth smallest character from a string

Question

I have a string input by the user and an integer k. How do I get the lexicographically kth smallest character from the input string?

Here is what I have been trying

static Character find(int K, String s){
    SortedSet<Character> ss = new TreeSet<Character>();
    for(Character c : s.toCharArray()){
        ss.add(c);
    }
    Iterator it = ss.iterator();
    int i=0;
      while (it.hasNext()) {
         Object element = it.next();
         if(i==K-1)
             return (Character) element;

      }
      return null;
}

make a sorted set of all characters in `str`. the rest should be obvious. — , Jul 10 '15 at 18:52
I am new to this. It'd be great if some one could help me with a function for the same. — g19, Jul 10 '15 at 18:53
@g19 Great, welcome! It would be better if you show us some code that you try and how you think it should be the way to make it work.... — melli-182, Jul 10 '15 at 18:55
*"Here is what i have been trying"* And what is your result? Btw: [What is a raw type and why shouldn't we use it?](http://stackoverflow.com/q/2770321) — Tom, Jul 10 '15 at 19:26
I think all you need to change is `if(i==K-1)` to `if(i++==K-1)` — laune, Jul 10 '15 at 19:29

score 1 · Accepted Answer · answered Jul 10 '15 at 20:32

Your code is OK, except for one thing: the handling of i variable. You should increment it at the end of the while loop in order to make it match K - 1.

There's also a more java8-friendly way of doing the same:

String str = "hello";
int k = 2;

Optional<Character> kth = 
    Stream.of(str.split(""))
        .map(c -> c.charAt(0))
        .sorted()
        .distinct()
        .skip(k - 1)
        .findFirst();

System.out.println(kth.isPresent() ? kth.get() : "k too big!"); // h

This code assumes that k is 1-based and that you're not taking duplicates into account to get the kth character.

Also, as pointed out in the comments, you shouldn't use a raw type for your Iterator. Consider changing it to Iterator<Character>, so you don't have to cast the value returned by it.next().

Lexicographically kth smallest character from a string

1 Answers1