Removing duplicates from a String in Java

Question

I am trying to iterate through a string in order to remove the duplicates characters.

For example the String aabbccdef should become abcdef and the String abcdabcd should become abcd

Here is what I have so far:

public class test {

    public static void main(String[] args) {

        String input = new String("abbc");
        String output = new String();

        for (int i = 0; i < input.length(); i++) {
            for (int j = 0; j < output.length(); j++) {
                if (input.charAt(i) != output.charAt(j)) {
                    output = output + input.charAt(i);
                }
            }
        }

        System.out.println(output);

    }

}

What is the best way to do this?

Answer 1

Convert the string to an array of char, and store it in a LinkedHashSet. That will preserve your ordering, and remove duplicates. Something like:

String string = "aabbccdefatafaz";

char[] chars = string.toCharArray();
Set<Character> charSet = new LinkedHashSet<Character>();
for (char c : chars) {
    charSet.add(c);
}

StringBuilder sb = new StringBuilder();
for (Character character : charSet) {
    sb.append(character);
}
System.out.println(sb.toString());

Answer 2

Using Stream makes it easy.

noDuplicates = Arrays.asList(myString.split(""))
                     .stream()
                     .distinct()
                     .collect(Collectors.joining());

Here is some more documentation about Stream and all you can do with it : https://docs.oracle.com/javase/8/docs/api/java/util/stream/package-summary.html

The 'description' part is very instructive about the benefits of Streams.

Answer 3

Try this simple solution:

public String removeDuplicates(String input){
    String result = "";
    for (int i = 0; i < input.length(); i++) {
        if(!result.contains(String.valueOf(input.charAt(i)))) {
            result += String.valueOf(input.charAt(i));
        }
    }
    return result;
}

Answer 4

I would use the help of LinkedHashSet. Removes dups (as we are using a Set, maintains the order as we are using linked list impl). This is kind of a dirty solution. there might be even a better way.

String s="aabbccdef";
Set<Character> set=new LinkedHashSet<Character>();
for(char c:s.toCharArray())
{
    set.add(Character.valueOf(c));
}

Answer 5

Here is an improvement to the answer by Dave.

It uses HashSet instead of the slightly more costly LinkedHashSet, and reuses the chars buffer for the result, eliminating the need for a StringBuilder.

String string = "aabbccdefatafaz";

char[] chars = string.toCharArray();
Set<Character> present = new HashSet<>();
int len = 0;
for (char c : chars)
    if (present.add(c))
        chars[len++] = c;

System.out.println(new String(chars, 0, len));   // abcdeftz

Answer 6

Create a StringWriter. Run through the original string using charAt(i) in a for loop. Maintain a variable of char type keeping the last charAt value. If you iterate and the charAt value equals what is stored in that variable, don't add to the StringWriter. Finally, use the StringWriter.toString() method and get a string, and do what you need with it.

Answer 7

Java 8 has a new String.chars() method which returns a stream of characters in the String. You can use stream operations to filter out the duplicate characters like so:

String out = in.chars()
            .mapToObj(c -> Character.valueOf((char) c)) // bit messy as chars() returns an IntStream, not a CharStream (which doesn't exist)
            .distinct()
            .map(Object::toString)
            .collect(Collectors.joining(""));

Answer 8

public class RemoveRepeated4rmString {

    public static void main(String[] args) {
        String s = "harikrishna";
        String s2 = "";
        for (int i = 0; i < s.length(); i++) {
            Boolean found = false;
            for (int j = 0; j < s2.length(); j++) {
                if (s.charAt(i) == s2.charAt(j)) {
                    found = true;
                    break; //don't need to iterate further
                }
            }
            if (found == false) {
                s2 = s2.concat(String.valueOf(s.charAt(i)));
            }
        }
        System.out.println(s2);
    }
}

Answer 9

    String input = "AAAB";

    String output = "";
    for (int index = 0; index < input.length(); index++) {
        if (input.charAt(index % input.length()) != input
                .charAt((index + 1) % input.length())) {

            output += input.charAt(index);

        }
    }
    System.out.println(output);

but you cant use it if the input has the same elements, or if its empty!

Answer 10

Code to remove the duplicate characters in a string without using any additional buffer. NOTE: One or two additional variables are fine. An extra array is not:

import java.util.*;
public class Main{
    public static char[] removeDupes(char[] arr){
        if (arr == null || arr.length < 2)
            return arr;
        int len = arr.length;
        int tail = 1;
        for(int x = 1; x < len; x++){
            int y;
            for(y = 0; y < tail; y++){
                if (arr[x] == arr[y]) break;
            }
            if (y == tail){
                arr[tail] = arr[x];
                tail++;
            }
        }
        return Arrays.copyOfRange(arr, 0, tail);
    }

    public static char[] bigArr(int len){
        char[] arr = new char[len];
        Random r = new Random();
        String alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890!@#$%^&*()-=_+[]{}|;:',.<>/?`~";

        for(int x = 0; x < len; x++){
            arr[x] = alphabet.charAt(r.nextInt(alphabet.length()));
        }

        return arr;
    }
    public static void main(String args[]){

        String result = new String(removeDupes(new char[]{'a', 'b', 'c', 'd', 'a'}));
        assert "abcd".equals(result) : "abcda should return abcd but it returns: " + result;

        result = new String(removeDupes(new char[]{'a', 'a', 'a', 'a'}));
        assert "a".equals(result) : "aaaa should return a but it returns: " + result;

        result = new String(removeDupes(new char[]{'a', 'b', 'c', 'a'}));
        assert "abc".equals(result) : "abca should return abc but it returns: " + result;

        result = new String(removeDupes(new char[]{'a', 'a', 'b', 'b'}));
        assert "ab".equals(result) : "aabb should return ab but it returns: " + result;

        result = new String(removeDupes(new char[]{'a'}));
        assert "a".equals(result) : "a should return a but it returns: " + result;

        result = new String(removeDupes(new char[]{'a', 'b', 'b', 'a'}));
        assert "ab".equals(result) : "abba should return ab but it returns: " + result;


        char[] arr = bigArr(5000000);
        long startTime = System.nanoTime();
        System.out.println("2: " + new String(removeDupes(arr)));
        long endTime = System.nanoTime();
        long duration = (endTime - startTime);
        System.out.println("Program took: " + duration + " nanoseconds");
        System.out.println("Program took: " + duration/1000000000 + " seconds");

    }
}

How to read and talk about the above code:

1. The method called removeDupes takes an array of primitive char called arr.
2. arr is returned as an array of primitive characters "by value". The arr passed in is garbage collected at the end of Main's member method removeDupes.
3. The runtime complexity of this algorithm is O(n) or more specifically O(n+(small constant)) the constant being the unique characters in the entire array of primitive chars.
4. The copyOfRange does not increase runtime complexity significantly since it only copies a small constant number of items. The char array called arr is not stepped all the way through.
5. If you pass null into removeDupes, the method returns null.
6. If you pass an empty array of primitive chars or an array containing one value, that unmodified array is returned.
7. Method removeDupes goes about as fast as physically possible, fully utilizing the L1 and L2 cache, so Branch redirects are kept to a minimum.
8. A 2015 standard issue unburdened computer should be able to complete this method with an primitive char array containing 500 million characters between 15 and 25 seconds.

Explain how this code works:

The first part of the array passed in is used as the repository for the unique characters that are ultimately returned. At the beginning of the function the answer is: "the characters between 0 and 1" as between 0 and tail.

We define the variable y outside of the loop because we want to find the first location where the array index that we are looking at has been duplicated in our repository. When a duplicate is found, it breaks out and quits, the y==tail returns false and the repository is not contributed to.

when the index x that we are peeking at is not represented in our repository, then we pull that one and add it to the end of our repository at index tail and increment tail.

At the end, we return the array between the points 0 and tail, which should be smaller or equal to in length to the original array.

Talking points exercise for coder interviews:

Will the program behave differently if you change the y++ to ++y? Why or why not.

Does the array copy at the end represent another 'N' pass through the entire array making runtime complexity O(n*n) instead of O(n) ? Why or why not.

Can you replace the double equals comparing primitive characters with a .equals? Why or why not?

Can this method be changed in order to do the replacements "by reference" instead of as it is now, "by value"? Why or why not?

Can you increase the efficiency of this algorithm by sorting the repository of unique values at the beginning of 'arr'? Under which circumstances would it be more efficient?

Answer 11

 public static void main(String a[]){
      String name="Madan";
      System.out.println(name);
      StringBuilder sb=new StringBuilder(name);
      for(int i=0;i<name.length();i++){
          for(int j=i+1;j<name.length();j++){
             if(name.charAt(i)==name.charAt(j)){
              sb.deleteCharAt(j);

             }
          }
      }
     System.out.println("After deletion :"+sb+"");

    }

Answer 12

import java.util.Scanner;

public class dublicate {
    public static void main(String... a) {
        System.out.print("Enter the String");
        Scanner Sc = new Scanner(System.in);
        String st=Sc.nextLine();
        StringBuilder sb=new StringBuilder();
        boolean [] bc=new boolean[256];
        for(int i=0;i<st.length();i++)
        {
            int index=st.charAt(i);
            if(bc[index]==false)
            {
                sb.append(st.charAt(i));
                bc[index]=true;
            }

        }
        System.out.print(sb.toString());
    }
}

Answer 13

To me it looks like everyone is trying way too hard to accomplish this task. All we are concerned about is that it copies 1 copy of each letter if it repeats. Then because we are only concerned if those characters repeat one after the other the nested loops become arbitrary as you can just simply compare position n to position n + 1. Then because this only copies things down when they're different, to solve for the last character you can either append white space to the end of the original string, or just get it to copy the last character of the string to your result.

String removeDuplicate(String s){

    String result = "";

    for (int i = 0; i < s.length(); i++){
        if (i + 1 < s.length() && s.charAt(i) != s.charAt(i+1)){
            result = result + s.charAt(i);
        }
        if (i + 1 == s.length()){
            result = result + s.charAt(i);
        }
    }

    return result;

}

Answer 14

String str1[] ="Hi helloo helloo  oooo this".split(" "); 

Set<String> charSet = new LinkedHashSet<String>();
for (String c: str1) 
{
       charSet.add(c); 
}
StringBuilder sb = new StringBuilder(); 
for (String character : charSet) 
{
       sb.append(character); 
}

System.out.println(sb.toString());

Answer 15

I think working this way would be more easy,,, Just pass a string to this function and the job is done :) .

private static void removeduplicate(String name)
{   char[] arr = name.toCharArray();
    StringBuffer modified =new StringBuffer();
    for(char a:arr)
    {
        if(!modified.contains(Character.toString(a)))
        {
            modified=modified.append(Character.toString(a)) ;
        }
    }
    System.out.println(modified);
}

Answer 16

public class RemoveDuplicatesFromStingsMethod1UsingLoops {

    public static void main(String[] args) {

        String input = new String("aaabbbcccddd");
        String output = "";
        for (int i = 0; i < input.length(); i++) {
            if (!output.contains(String.valueOf(input.charAt(i)))) {
                output += String.valueOf(input.charAt(i));
            }
        }
        System.out.println(output);
    }
}

output: abcd

Answer 17

public static void alpha(char[] finalname)
{
    if (finalname == null)
    {
        return;
    }
    
    if (finalname.length <2)
    {
        return;
    }
    
    char empty = '\000';
    for (int i=0; i<finalname.length-1; i++)
    {
        if (finalname[i] == finalname[i+1])
        {
            finalname[i] = empty;
        }
    }
    
    String alphaname = String.valueOf(finalname);
    alphaname = alphaname.replace("\000", "");
    System.out.println(alphaname);
    
        
}

Answer 18

Oldschool way (as we wrote such a tasks in Apple ][ Basic, adapted to Java):

int i,j;
StringBuffer str=new StringBuffer();
Scanner in = new Scanner(System.in);
System.out.print("Enter string: ");
str.append(in.nextLine());

for (i=0;i<str.length()-1;i++){
    for (j=i+1;j<str.length();j++){
        if (str.charAt(i)==str.charAt(j))
            str.deleteCharAt(j);
    }
}
System.out.println("Removed non-unique symbols: " + str);

Answer 19

Here is another logic I'd like to share. You start comparing from midway of the string length and go backward.

Test with: input = "azxxzy"; output = "ay";

String removeMidway(String input){
        cnt = cnt+1;
        StringBuilder str = new StringBuilder(input);
        int midlen = str.length()/2;
        for(int i=midlen-1;i>0;i--){

            for(int j=midlen;j<str.length()-1;j++){     
                if(str.charAt(i)==str.charAt(j)){
                    str.delete(i, j+1);
                    midlen = str.length()/2;
                    System.out.println("i="+i+",j="+j+ ",len="+ str.length() + ",midlen=" + midlen+ ", after deleted = " + str);
                }
            }
        }       
        return str.toString();
    }

Answer 20

void remove_duplicate (char* str, int len) {
    unsigned int index = 0;
    int c = 0;
    int i = 0;
    while (c < len) {
        /* this is just example more check can be added for
           capital letter, space and special chars */

        int pos = str[c] - 'a';
        if ((index & (1<<pos)) == 0) {
            str[i++] = str[c];
            index |= (1<<pos);
        }
        c++;
    }
    str[i] = 0;
}

Answer 21

Another possible solution, in case a string is an ASCII string, is to maintain an array of 256 boolean elements to denote ASCII character appearance in a string. If a character appeared for the first time, we keep it and append to the result. Otherwise just skip it.

public String removeDuplicates(String input) {
    boolean[] chars = new boolean[256];
    StringBuilder resultStringBuilder = new StringBuilder();
    for (Character c : input.toCharArray()) {
        if (!chars[c]) {
            resultStringBuilder.append(c);
            chars[c] = true;
        }
    }
    return resultStringBuilder.toString();
}

This approach will also work with Unicode string. You just need to increase chars size.

Answer 22

Solution using JDK7:

public static String removeDuplicateChars(final String str){

    if (str == null || str.isEmpty()){
        return str;
    }

    final char[] chArray = str.toCharArray();
    final Set<Character> set = new LinkedHashSet<>();
    for (char c : chArray) {
        set.add(c);
    }

    final StringBuilder sb = new StringBuilder();
    for (Character character : set) {
        sb.append(character);
    }
    return sb.toString();
}

Answer 23

    String str = "eamparuthik@gmail.com";
    char[] c = str.toCharArray();
    String op = "";

    for(int i=0; i<=c.length-1; i++){
        if(!op.contains(c[i] + ""))
        op = op + c[i];
    }
    System.out.println(op);

Answer 24

public static String removeDuplicateChar(String str){
         char charArray[] = str.toCharArray();
         StringBuilder stringBuilder= new StringBuilder();
         for(int i=0;i<charArray.length;i++){
             int index = stringBuilder.toString().indexOf(charArray[i]);
             if(index <= -1){
                 stringBuilder.append(charArray[i]);
             }
         }
         return stringBuilder.toString();
    }

Answer 25

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class RemoveDuplicacy
{
        public static void main(String args[])throws IOException
        {
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            System.out.print("Enter any word : ");
            String s = br.readLine();
            int l = s.length();
            char ch;
            String ans=" ";

            for(int i=0; i<l; i++)
            {
                ch = s.charAt(i);
                if(ch!=' ')
                    ans = ans + ch;
                s = s.replace(ch,' '); //Replacing all occurrence of the current character by a space
            }

           System.out.println("Word after removing duplicate characters : " + ans);
        }

}

Answer 26

public static void main(String[] args) {

    int i,j;
    StringBuffer str=new StringBuffer();
    Scanner in = new Scanner(System.in);
    System.out.print("Enter string: ");

    str.append(in.nextLine());

    for (i=0;i<str.length()-1;i++)
    {
        for (j=1;j<str.length();j++)
        {
            if (str.charAt(i)==str.charAt(j))
                str.deleteCharAt(j);
        }
    }
    System.out.println("Removed String: " + str);
}

Answer 27

This is improvement on solution suggested by @Dave. Here, I am implementing in single loop only.

Let's reuse the return of set.add(T item) method and add it simultaneously in StringBuffer if add is successfull

This is just O(n). No need to make a loop again.

String string = "aabbccdefatafaz";

char[] chars = string.toCharArray();
StringBuilder sb = new StringBuilder();
Set<Character> charSet = new LinkedHashSet<Character>();
for (char c : chars) {
    if(charSet.add(c) ){
        sb.append(c);
    }

}
System.out.println(sb.toString()); // abcdeftz

Answer 28

Simple solution is to iterate through the given string and put each unique character into another string(in this case, a variable result ) if this string doesn't contain that particular character.Finally return result string as output.

Below is working and tested code snippet for removing duplicate characters from the given string which has O(n) time complexity .

private static String removeDuplicate(String s) {
      String result="";
      for (int i=0 ;i<s.length();i++) {
          char ch = s.charAt(i);
          if (!result.contains(""+ch)) {
              result+=""+ch;
          }
      }
      return result;
  }

If the input is madam then output will be mad.
If the input is anagram then output will be angrm

Hope this helps.
Thanks

Answer 29

For the simplicity of the code- I have taken hardcore input, one can take input by using Scanner class also

    public class KillDuplicateCharInString {
    public static void main(String args[]) {
        String str= "aaaabccdde ";
        char arr[]= str.toCharArray();
        int n = arr.length;
        String finalStr="";
        for(int i=0;i<n;i++) {
            if(i==n-1){
                finalStr+=arr[i];
                break;
            }
            if(arr[i]==arr[i+1]) {
                continue;
            }
            else {
                finalStr+=arr[i];
            }
        }
        System.out.println(finalStr);



    }
}

Answer 30

 public static void main (String[] args)
 {
    Scanner sc = new Scanner(System.in);
    String s = sc.next();
    String str = "";
    char c;
    for(int i = 0; i < s.length(); i++)
    {
        c = s.charAt(i);
        str = str + c;
        s = s.replace(c, ' ');
        if(i == s.length() - 1)
        {
           System.out.println(str.replaceAll("\\s", ""));   
        }
    }
}

Answer 31

package com.st.removeduplicate;
 public class RemoveDuplicate {
   public static void main(String[] args) {
    String str1="shushil",str2="";      
    for(int i=0; i<=str1.length()-1;i++) {
        int count=0;
        for(int j=0;j<=i;j++) {
            if(str1.charAt(i)==str1.charAt(j)) 
                count++;
            if(count >1)
                break;
        }
        if(count==1) 
            str2=str2+str1.charAt(i);
    }
    System.out.println(str2);

}

}

Answer 32

You can't. You can create a new String that has duplicates removed. Why aren't you using StringBuilder (or StringBuffer, presumably)?

You can run through the string and store the unique characters in a char[] array, keeping track of how many unique characters you've seen. Then you can create a new String using the String(char[], int, int) constructor.

Also, the problem is a little ambiguous—does “duplicates” mean adjacent repetitions? (In other words, what should happen with abcab?)

Answer 33

Hope this will help.

public void RemoveDuplicates() {
    String s = "Hello World!";
    int l = s.length();
    char ch;
    String result = "";
    for (int i = 0; i < l; i++) {
        ch = s.charAt(i);
        if (ch != ' ') {
            result = result + ch;
        }
        // Replacing space in all occurrence of the current character
        s = s.replace(ch, ' ');
    }
    System.out.println("After removing duplicate characters : " + result);
}

Answer 34

've an array to know whether a character has already been recorded or not; if not, add that to the string buffer. Please note that I have made it case sensitive; with the int array, you can always make it('ve not done that in this code) to return the number of occurrences too.

private static String removeDuplicates(String s) {

    int [] occurrences = new int[52];
    Arrays.fill(occurrences,0);

    StringBuffer deDupS = new StringBuffer();
    for(int i = 0; i < s.length(); i++) {
        if(s.charAt(i) >= 97) {
            if(occurrences[s.charAt(i) - 97] == 0) {
                deDupS.append(s.charAt(i));
                occurrences[s.charAt(i) - 97]++;
            }
        } else if(s.charAt(i) >= 65) {
            if(occurrences[s.charAt(i) - 65 + 26] == 0) {
                deDupS.append(s.charAt(i));
                occurrences[s.charAt(i) - 65 + 26]++;
            }
        }
    }

    return deDupS.toString();

}

Answer 35

StringBuilder builderWord = new StringBuilder(word);
 for(int index=0; index < builderWord.length(); index++) {
   for(int reverseIndex=builderWord.length()-1; reverseIndex > index;reverseIndex--) {
     if (builderWord.charAt(reverseIndex) == builderWord.charAt(index)) {
       builderWord.deleteCharAt(reverseIndex);
     }
   }
}
return builderWord.toString();

Answer 36

public String removeDuplicates(String dupCharsString){
    StringBuffer buffer = new StringBuffer(dupCharsString);
    int step = 0;
    while(step <= buffer.length()){
        for( int i = step + 1; i < buffer.length(); i++ ){
            if( buffer.charAt(i) == buffer.charAt(step) ){
                buffer.setCharAt(i, ' ');
            }
        }
        step++;
    }
    return buffer.toString().replaceAll("\\s","");
}

Answer 37

import java.util.LinkedHashMap;
import java.util.Map.Entry;

public class Sol {

    public static void main(String[] args) {
        char[] str = "bananas".toCharArray();
        LinkedHashMap<Character,Integer> map = new LinkedHashMap<>();
        StringBuffer s = new StringBuffer();

        for(Character c : str){
            if(map.containsKey(c))
                map.put(c, map.get(c)+1);
            else
                map.put(c, 1);
        }

        for(Entry<Character,Integer> entry : map.entrySet()){
            s.append(entry.getKey());
        }

        System.out.println(s);
    }

}

Answer 38

String s = "WelcomeToJava";
String result = "";
List<String> al = new ArrayList<String>();

for (int i = 0; i < s.length(); i++) {
    for (int k = 0; k < s.length(); k++) {
        if (String.valueOf(s.charAt(i)).equalsIgnoreCase(String.valueOf(s.charAt(k)))) {
            if (!al.contains(String.valueOf(s.charAt(i)))) {
                al.add(String.valueOf(s.charAt(i)));
                result = result + String.valueOf(s.charAt(i));
            }
        }
    }
}

System.out.println("Result--" + result);

Answer 39

Arrays.stream(input.split("")).distinct().collect(joining());

Answer 40

 String input = "abbcccaabbddeessacccbbddefgaabbccddeeffggadscsda";
    String output ="";

    for (int i= 0; i<input.length();i++) {
        if (!output.contains(input.substring(i, i+1))) 
            output = output+ input.substring(i, i+1);

    }


        System.out.println(output);

Answer 41

package StringPrograms;

public class RemoveDuplicateCharacters {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        boolean flag;
        String str = "Stackoverflowtest";
        String a = "";
        int dlen = a.length();
        for (int i = 0; i < str.length(); i++) {
            flag = false;
            for (int j = 0; j <dlen; j++)
                
                 if (str.charAt(i) == a.charAt(j)) {
                    flag = true;
                    break;
                }
            if (flag == false) {
                a = a + str.charAt(i);
                dlen = dlen + 1;
            }
        }
        System.out.println(a);

    }

}

Answer 42

String output = "";
    HashMap<Character,Boolean> map = new HashMap<>();
    for(int i=0;i<str.length();i++){
        char ch = str.charAt(i);
        if (!map.containsKey(ch)){
            output += ch;
            map.put(ch,true);
        }
    }
    return output;

This is the simple approach using HashMap in Java in one pass .

Time Complexity : O(n)

Space Complexity : O(n)

Answer 43

package JavaPracticePackage;

public class Replace_same_char {

    public static void main(String[] args) {
        String value = "AABBCCDD";
        String FinalValue = "";
        for(int i=0; i<value.length();i++) {
            if(FinalValue.indexOf(value.charAt(i))==-1) {
                FinalValue=FinalValue+value.charAt(i);
            }
        }
        System.out.println(FinalValue);
    }
}

Answer 44

You can use HashSet class in java:

public static String removeDuplicates(String str) {
        char[] chars = str.toCharArray();
        Set<String> set = new HashSet<>();
        for (int i = 0; i < chars.length; i++) {
            set.add(String.valueOf(chars[i]));
        }
       return String.join("", set);
    }

Example:

public class Test {
    public static void main(String[] args) {
        System.out.println(removeDuplicates("aabbccdd")); // abcd
    }
}

Answer 45

Try this simple solution using Set collection concept: String str = "aabbcdegg";

    Set<Character>removeduplicates = new LinkedHashSet<>();
    char strarray[]= str.toCharArray();
    for(char c:strarray)
    {
        removeduplicates.add(c);
    }


    Iterator<Character> itr = removeduplicates.iterator();
    while(itr.hasNext())
    {
        System.out.print(itr.next());
    }

Answer 46

public class StringTest {

    public static String dupRemove(String str) {

        Set<Character> s1 = new HashSet<Character>();
        StringBuffer sb = new StringBuffer();
        for (int i = 0; i < str.length(); i++) {

            Character c = str.charAt(i);

            if (!s1.contains(c)) {
                s1.add(c);
                sb.append(c);
            }


        }

        System.out.println(sb.toString());

        return sb.toString();

    }

    public static void main(String[] args) {

        dupRemove("AAAAAAAAAAAAAAAAA BBBBBBBB");
    }

}

Answer 47

  val str="karunkumardev"
  val arr=str.toCharArray()
  val len=arr.size-1

  for(i in 0..len){
      var flag=false
      for(j in i+1..len){
          if(arr[i]==arr[j]){
             flag=true
              break
          }
      }
      if(!flag){
          print(arr[i])
      }
  }

Answer 48

public class RemoveDuplicate {

    static String st="papuuu";
    static String st1="";
    public static void main(String[] args) {
        for (int i = 0; i < st.length(); i++) {
            String ff = String.valueOf(st.charAt(i));
            if (!st1.contains(ff)) {
                st1 = st1 + st.charAt(i);
            }
        }   
        System.out.println(st1);
    }
}

Answer 49

  let str = 'abcdabddbbccedbd'; 
  let res = "";
  new Set([...str]).forEach((val) => {
      res += val
  })

  console.log(res);

Answer 50

package com.core.interview.client;

import java.util.LinkedHashSet;

import java.util.Scanner;

import java.util.Set;

public class RemoveDuplicateFromString {

    public static String DupRemoveFromString(String str) {

        char[] c1 = str.toCharArray();

        Set < Character > charSet = new LinkedHashSet < Character > ();

        for (char c: c1) {

            charSet.add(c);
        }

        StringBuffer sb = new StringBuffer();

        for (Character c2: charSet) {

            sb.append(c2);
        }

        return sb.toString();

    }

    public static void main(String[] args) {

        System.out.println("Enter Your String: ");

        Scanner sc = new Scanner(System. in );

        String str = sc.nextLine();

        System.out.println(DupRemoveFromString(str));
    }
}