Java: Print a unique character in a string

Question

I\'m writing a program that will print the unique character in a string (entered through a scanner). I\'ve created a method that tries to accomplish this but I keep getting

Answer 1

Based on your desired output you can replace each character already present with a blank character.

public static void uniqueCharacters(String test){
  String temp = "";
  for(int i = 0; i < test.length(); i++){
      if (temp.indexOf(test.charAt(i)) == - 1){
         temp = temp + test.charAt(i);
      } else {
         temp.replace(String.valueOf(temp.charAt(i)), "");
      }
 }

System.out.println(temp + " ");

}

Answer 2

If you don't want to use additional space:

    String abc="developer";

    System.out.println("The unique characters are-");

    for(int i=0;i<abc.length();i++)
    {
        for(int j=i+1;j<abc.length();j++)
        {
            if(abc.charAt(i)==abc.charAt(j))
                abc=abc.replace(String.valueOf(abc.charAt(j))," ");
        }
    }   
    System.out.println(abc);

Time complexity O(n^2) and no space.

Answer 3

Though to approach a solution I would suggest you to try and use a better data structure and not just string. Yet, you can simply modify your logic to delete already existing duplicates using an else as follows :

public static void uniqueCharacters(String test) {
        String temp = "";
        for (int i = 0; i < test.length(); i++) {
            char ch = test.charAt(i);
            if (temp.indexOf(ch) == -1) {
                temp = temp + ch;
            } else {
                temp.replace(String.valueOf(ch),""); // added this to your existing code
            }
        }

        System.out.println(temp + " ");

    }

Answer 4

Based on your desired output, you have to replace a character that initially has been already added when it has a duplicated later, so:

public static void uniqueCharacters(String test){
    String temp = "";
    for (int i = 0; i < test.length(); i++){
        char current = test.charAt(i);
        if (temp.indexOf(current) < 0){
            temp = temp + current;
        } else {
            temp = temp.replace(String.valueOf(current), "");
        }
    }

    System.out.println(temp + " ");

}

Answer 5

public class UniqueCharactersInString {


 public static void main(String []args){

    String input = "aabbcc";
    String output = uniqueString(input);

    System.out.println(output);
 }

 public static String uniqueString(String s){
     HashSet<Character> uniques = new HashSet<>();
     uniques.add(s.charAt(0));
     String out = "";
     out += s.charAt(0);

     for(int i =1; i < s.length(); i++){
         if(!uniques.contains(s.charAt(i))){
             uniques.add(s.charAt(i));
             out += s.charAt(i);
         }
     }
     return out;
 }
}

What would be the inneficiencies of this answer? How does it compare to other answers?

Answer 6

The accepted answer will not pass all the test case for example

input -"aaabcdd"

desired output-"bc"
but the accepted answer will give -abc

because the character a present odd number of times.

Here I have used ConcurrentHasMap to store character and the number of occurrences of character then removed the character if the occurrences is more than one time.

import java.util.concurrent.ConcurrentHashMap;

public class RemoveConductive {

    public static void main(String[] args) {

        String s="aabcddkkbghff";

        String[] cvrtar=s.trim().split("");

        ConcurrentHashMap<String,Integer> hm=new ConcurrentHashMap<>();
        for(int i=0;i<cvrtar.length;i++){
            if(!hm.containsKey(cvrtar[i])){
                hm.put(cvrtar[i],1);
            }
            else{
                 hm.put(cvrtar[i],hm.get(cvrtar[i])+1);
            }
        }
        for(String ele:hm.keySet()){
            if(hm.get(ele)>1){
                hm.remove(ele);
            }
        }
        for(String key:hm.keySet()){
            System.out.print(key);
        }
    }  
}