Counting unique characters in a String given by the user

Question

I have to write a program that counts the uniques characters in a String given by the user. For example \"abc\" returns 3 and \"aabbccd\" returns 4. I am not allow to use advan

Answer 1

Unique Characters In A String:

This is a basic Java interview subject, where the interviewer wants to check the knowledge of HashSet or indexOf(in case of Java 7) . Let's take up a question. Let's say the interview tells you to
Check if the String is unique : HashSet ensures uniqueness ,In other words , every object in the HashSet presents only once. Hence, we will use HashSet.

import java.util.HashSet;
import java.util.Set;

public class Abc {
public static void main(String[] args) {
    String a = "Gini";
    String aa = a.toLowerCase();
       if(  isUnique(aa)   ) {
              System.out.println("All characters are unique");
       }else {
           System.out.println("All characters  are not unique");
       }
}
public static boolean isUnique(String a ) {
    Set< Character> set = new HashSet<>();
    char[] charArray =a.toCharArray();
    for(Character ch :charArray) {
        if(!set.add(ch)) {
            return false;
        }//if
    }//foreach
        return true;
}
}

The output in case of GINI will be : All characters are not unique

Now, the count of the unique characters. Here also we will use HashSet because of the uniqueness.

import java.util.HashSet;

public class practice11 {
public static void main(String[] args) {
    String a = "Gini";
    String aa = a.toLowerCase();
    System.out.println(countUniqueCharacters(aa));  
}
public static int  countUniqueCharacters(String a) {
    char[] charArray = a.toCharArray();

    HashSet set = new HashSet();

    for(int i = 0 ; i< charArray.length ; i++) {

         set.add(charArray[i]);
    }//for
    return   set.size() ;//This will give 3
}
}

The output for Gini will be 3 (Gin will be considered) .

indexOf Method : indexOf() returns the index of first occurrence of the character and then we are comparing with -1.

public class Abc {
public static void main(String[] args) {
    String a = "Gini";
    String aa = a.toLowerCase();
    String t = " ";
    for (int i = 0; i < aa.length(); i++) {
            int  pp =   aa.charAt(i) ;

        if(t.indexOf(aa.charAt(i))  ==  -1  ) {
            t = t + aa.charAt(i);
        }//if


    }//for

    System.out.println(t );// This will give => gin
    System.out.println(t.length()); // this will give 3
}//main
}//end

In Java 8, this is extremely easy. We just have to use chars().distinct().count(). But the return type will be long.

class Abc{
    public static void main(String[] args) {
        String a = "Gini";
        String aa = a.toLowerCase();
        System.out.println(  countUnique(aa));
    }

    private static long  countUnique(String aa) {
        // this will give 3(gin. another i will be not be counted as we have used distinct())
         return  aa.chars().distinct().count() ; 
    }
}

Another classic Interview Question : Find the First non repeating character in the String OR the First Unique Character in a String. This problem can be solved using the knowledge of HashMap.

class Abc{
    public static void main(String[] args) {
          String a = "GinaRani" ;
// Output will be G
          System.out.println( firstNonRepeatingCharacter(a) );

    }//main
    public static Character firstNonRepeatingCharacter(String a){
      Map map = new HashMap<>();
        char[] charArray = a.toCharArray();
        for(    Character  ch   :     charArray){
               if( map.containsKey(ch)   )   {
                    map.put(ch, map.get(ch) +1 ) ;
               } else{
                   map.put(ch, 1);
               }
        }//for

        // 1st non repeating character

        for( int  i = 0 ; i < a.length(); i ++ ){
             char chh = a.charAt(i);
             if(  map.get(chh) == 1 ){
                 System.out.println("first non repeating character in the String is  : ");
                   return chh ; 
             }//if
        }//for

        return null; 

    }//firstNonRepeatingCharacter
}//end

We will do this with the help of list comprehension in Python. We are not using set() operator. In Python :

string = 'GiniGinaProtijayi'

unique = []

[ unique.append(ch)  for ch in string if ch not in unique ]
lengthofUniqueCharacters = len(unique)
print("length of Unique Characters => " ,lengthofUniqueCharacters)
print("as a list => ",unique)
print("as a string => " , ''.join(unique))

Just to print out the distinct characters in Java 8:

public class Test5 {
    public static void main(String[] args) {
        String a = "GinaGini";

        String aa = a.chars().distinct()
                .collect(StringBuilder::new, 
                        StringBuilder::appendCodePoint, 
                        StringBuilder::append)
                .toString();
        System.out.println(aa);//Gina

    }// main

}