Remove duplicates from a given string

Given a string S, the task is to remove all the duplicates in the given string. Below are the different methods to remove duplicates in a string.

METHOD 1 (Simple)

C++

// CPP program to remove duplicate character
// from character array and print in sorted
// order
#include <bits/stdc++.h>
using namespace std;

char *removeDuplicate(char str[], int n)
{
   // Used as index in the modified string
   int index = 0;

   // Traverse through all characters
   for (int i=0; i<n; i++) {

     // Check if str[i] is present before it
     int j;
     for (j=0; j<i; j++)
        if (str[i] == str[j])
           break;

     // If not present, then add it to
     // result.
     if (j == i)
        str[index++] = str[i];
   }

   return str;
}

// Driver code
int main()
{
   char str[]= "geeksforgeeks";
   int n = sizeof(str) / sizeof(str[0]);
   cout << removeDuplicate(str, n);
   return 0;
}

C

#include <stdio.h>
#include <string.h>

char* removeDuplicate(char str[], int n)
{
    // Used as an index in the modified string
    int index = 0;

    // Traverse through all characters
    for (int i = 0; i < n; i++) {
        // Check if str[i] is present before it
        int j;
        for (j = 0; j < i; j++) {
            if (str[i] == str[j])
                break;
        }

        // If not present, then add it to the result.
        if (j == i)
            str[index++] = str[i];
    }

    // Add null character at the end to terminate the string
    str[index] = '\0';

    return str;
}

// Driver code
int main()
{
    char str[] = "geeksforgeeks";
    int n = sizeof(str) / sizeof(str[0]);
    printf("%s\n", removeDuplicate(str, n));
    return 0;
}

Java

// Java program to remove duplicate character
// from character array and print in sorted
// order
import java.util.*;

class GFG
{
    static String removeDuplicate(char str[], int n)
    {
        // Used as index in the modified string
        int index = 0;

        // Traverse through all characters
        for (int i = 0; i < n; i++)
        {

            // Check if str[i] is present before it
            int j;
            for (j = 0; j < i; j++)
            {
                if (str[i] == str[j])
                {
                    break;
                }
            }

            // If not present, then add it to
            // result.
            if (j == i)
            {
                str[index++] = str[i];
            }
        }
        return String.valueOf(Arrays.copyOf(str, index));
    }

    // Driver code
    public static void main(String[] args)
    {
        char str[] = "geeksforgeeks".toCharArray();
        int n = str.length;
        System.out.println(removeDuplicate(str, n));
    }
}

// This code is contributed by Rajput-Ji

Python3

string="geeksforgeeks"
p=""
for char in string:
    if char not in p:
        p=p+char
print(p)
k=list("geeksforgeeks")

C#

// C# program to remove duplicate character
// from character array and print in sorted
// order
using System;
using System.Collections.Generic;
class GFG
{
static String removeDuplicate(char []str, int n)
{
    // Used as index in the modified string
    int index = 0;

    // Traverse through all characters
    for (int i = 0; i < n; i++)
    {

        // Check if str[i] is present before it
        int j;
        for (j = 0; j < i; j++)
        {
            if (str[i] == str[j])
            {
                break;
            }
        }

        // If not present, then add it to
        // result.
        if (j == i)
        {
            str[index++] = str[i];
        }
    }
    char [] ans = new char[index];
    Array.Copy(str, ans, index);
    return String.Join("", ans);
}

// Driver code
public static void Main(String[] args)
{
    char []str = "geeksforgeeks".ToCharArray();
    int n = str.Length;
    Console.WriteLine(removeDuplicate(str, n));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// JavaScript program to remove duplicate character
// from character array and print in sorted
// order
function removeDuplicate(str, n)
    {
        // Used as index in the modified string
        var index = 0;

        // Traverse through all characters
        for (var i = 0; i < n; i++)
        {

            // Check if str[i] is present before it
            var j;
            for (j = 0; j < i; j++)
            {
                if (str[i] == str[j])
                {
                    break;
                }
            }

            // If not present, then add it to
            // result.
            if (j == i)
            {
                str[index++] = str[i];
            }
        }

        return str.join("").slice(str, index);
    }

    // Driver code
        var str = "geeksforgeeks".split("");
        var n = str.length;
        document.write(removeDuplicate(str, n));

// This code is contributed by shivanisinghss2110

</script>

Output

geksfor

Time Complexity: O(n * n) 
Auxiliary Space: O(1), Keeps the order of elements the same as the input. 

METHOD 2 (using set)
Use set to store only one instance of any value. 

C++

// CPP program to remove duplicate character
// from character array and print in sorted
// order
#include <bits/stdc++.h>
using namespace std;

char *removeDuplicate(char str[], int n)
{
    // create a set using string characters
    // excluding '\0'
    unordered_set<char>s (str, str+n-1);

    // print content of the set
    int i = 0;
    for (auto x : s)
       str[i++] = x;
    str[i] = '\0';

    return str;
}

// Driver code
int main()
{
   char str[]= "geeksforgeeks";
   int n = sizeof(str) / sizeof(str[0]);
   cout << removeDuplicate(str, n);
   return 0;
}

Java

// Java program to remove duplicate character
// from character array and print in sorted
// order
import java.util.*;

class GFG {

    static void removeDuplicate(char str[], int n)
    {
       // Create a set using String characters
    // excluding '\0'
        HashSet<Character> s = new LinkedHashSet<>(n - 1);
      // HashSet doesn't allow repetition of elements
        for (char x : str)
            s.add(x);

        // Print content of the set
        for (char x : s)
            System.out.print(x);
    }

    // Driver code
    public static void main(String[] args)
    {
        char str[] = "geeksforgeeks".toCharArray();
        int n = str.length;

        removeDuplicate(str, n);
    }
}

// This code is contributed by todaysgaurav

Python3

# Python program to remove duplicate character
# from character array and print in sorted
# order
def removeDuplicate(str, n):
    s = set()

    # Create a set using String characters
    for i in str:
        s.add(i)

    # Print content of the set
    st = ""
    for i in s:
        st = st+i
    return st


# Driver code
str = "geeksforgeeks"
n = len(str)
print(removeDuplicate(list(str), n))

# This code is contributed by rajsanghavi9.

C#

// C# program to remove duplicate character
// from character array and print in sorted
// order
using System;
using System.Collections.Generic;


public class GFG{

static char []removeDuplicate(char []str, int n)
{

    // Create a set using String characters
    // excluding '\0'
    HashSet<char>s = new HashSet<char>(n - 1);
    foreach(char x in str)
        s.Add(x);

    char[] st = new char[s.Count];

    // Print content of the set
    int i = 0;
    foreach(char x in s)
       st[i++] = x;

    return st;
}

// Driver code
public static void Main(String[] args)
{
   char []str= "geeksforgeeks".ToCharArray();
   int n = str.Length;

   Console.Write(removeDuplicate(str, n));
}
}

// This code contributed by gauravrajput1

Javascript

<script>
// javascript program to remove duplicate character
// from character array and print in sorted
// order

    function removeDuplicate( str , n)
    {

        // Create a set using String characters
        // excluding '\0'
        var s = new Set();

        // HashSet doesn't allow repetition of elements
        for (var i = 0;i<n;i++)
            s.add(str[i]);

        // Print content of the set
        for (const v of s) {

            document.write(v);
    }
    }

    // Driver code
        var str = "geeksforgeeks";
        var n = str.length;

        removeDuplicate(str, n);

// This code is contributed by umadevi9616
</script>

Output

ofskreg

Time Complexity: O(n) 
Auxiliary Space: O(n)

Thanks to Anivesh Tiwari for suggesting this approach.

It does not keep the order of elements the same as the input but prints them in sorted order.

METHOD 3 (Use Sorting)
Algorithm:

  1) Sort the elements.
  2) Now in a loop, remove duplicates by comparing the 
      current character with previous character.
  3)  Remove extra characters at the end of the resultant string.

Example:

Input string:  geeksforgeeks
1) Sort the characters
   eeeefggkkorss
2) Remove duplicates
    efgkorskkorss
3) Remove extra characters
     efgkors

Note that, this method doesn’t keep the original order of the input string. For example, if we are to remove duplicates for geeksforgeeks and keep the order of characters the same, then the output should be geksfor, but the above function returns efgkos. We can modify this method by storing the original order.

Implementation:

C++

// C++ program to remove duplicates, the order of
// characters is not maintained in this program
#include<bits/stdc++.h>
using namespace std;

/* Function to remove duplicates in a sorted array */
char *removeDupsSorted(char *str)
{
    int res_ind = 1, ip_ind = 1;

    /* In place removal of duplicate characters*/
    while (*(str + ip_ind))
    {
        if (*(str + ip_ind) != *(str + ip_ind - 1))
        {
            *(str + res_ind) = *(str + ip_ind);
            res_ind++;
        }
        ip_ind++;
    }

    /* After above step string is efgkorskkorss.
       Removing extra kkorss after string*/
    *(str + res_ind) = '\0';

    return str;
}

/* Function removes duplicate characters from the string
   This function work in-place and fills null characters
   in the extra space left */
char *removeDups(char *str)
{
   int n = strlen(str);

   // Sort the character array
   sort(str, str+n);

   // Remove duplicates from sorted
   return removeDupsSorted(str);
}

/* Driver program to test removeDups */
int main()
{
  char str[] = "geeksforgeeks";
  cout << removeDups(str);
  return 0;
}

C

// C program to remove duplicates, the order of
// characters is not maintained in this program
# include <stdio.h>
# include <stdlib.h>
# include <string.h>
/* Function to remove duplicates in a sorted array */
char *removeDupsSorted(char *str);

/* Utility function to sort array A[] */
void quickSort(char A[], int si, int ei);

/* Function removes duplicate characters from the string
   This function work in-place and fills null characters
   in the extra space left */
char *removeDups(char *str)
{
  int len = strlen(str);
  quickSort(str, 0, len-1);
  return removeDupsSorted(str);
}

/* Function to remove duplicates in a sorted array */
char *removeDupsSorted(char *str)
{
  int res_ind = 1, ip_ind = 1;

  /* In place removal of duplicate characters*/
  while (*(str + ip_ind))
  {
    if (*(str + ip_ind) != *(str + ip_ind - 1))
    {
      *(str + res_ind) = *(str + ip_ind);
      res_ind++;
    }
    ip_ind++;
  }

  /* After above step string is efgkorskkorss.
     Removing extra kkorss after string*/
  *(str + res_ind) = '\0';

  return str;
}

/* Driver program to test removeDups */
int main()
{
  char str[] = "geeksforgeeks";
  printf("%s", removeDups(str));
  getchar();
  return 0;
}

/* FOLLOWING FUNCTIONS ARE ONLY FOR SORTING
    PURPOSE */
void exchange(char *a, char *b)
{
  char temp;
  temp = *a;
  *a   = *b;
  *b   = temp;
}

int partition(char A[], int si, int ei)
{
  char x = A[ei];
  int i = (si - 1);
  int j;

  for (j = si; j <= ei - 1; j++)
  {
    if (A[j] <= x)
    {
      i++;
      exchange(&A[i], &A[j]);
    }
  }
  exchange (&A[i + 1], &A[ei]);
  return (i + 1);
}

/* Implementation of Quick Sort
A[] --> Array to be sorted
si  --> Starting index
ei  --> Ending index
*/
void quickSort(char A[], int si, int ei)
{
  int pi;    /* Partitioning index */
  if (si < ei)
  {
    pi = partition(A, si, ei);
    quickSort(A, si, pi - 1);
    quickSort(A, pi + 1, ei);
  }
}

Java

// Java program to remove duplicates, the order of
// characters is not maintained in this program

import java.util.Arrays;

public class GFG
{
    /* Method to remove duplicates in a sorted array */
    static String removeDupsSorted(String str)
    {
        int res_ind = 1, ip_ind = 1;

        // Character array for removal of duplicate characters
        char arr[] = str.toCharArray();

        /* In place removal of duplicate characters*/
        while (ip_ind != arr.length)
        {
            if(arr[ip_ind] != arr[ip_ind-1])
            {
                arr[res_ind] = arr[ip_ind];
                res_ind++;
            }
            ip_ind++;

        }

        str = new String(arr);
        return str.substring(0,res_ind);
    }

    /* Method removes duplicate characters from the string
       This function work in-place and fills null characters
       in the extra space left */
    static String removeDups(String str)
    {
       // Sort the character array
       char temp[] = str.toCharArray();
       Arrays.sort(temp);
       str = new String(temp);

       // Remove duplicates from sorted
       return removeDupsSorted(str);
    }

    // Driver Method
    public static void main(String[] args)
    {
        String str = "geeksforgeeks";
        System.out.println(removeDups(str));
    }
}

Python3

# Python3 program to remove duplicates, the order of
# characters is not maintained in this program

# Utility function to convert string to list
def toMutable(string):
    temp = []
    for x in string:
        temp.append(x)
    return temp

# Utility function to convert string to list
def toString(List):
    return ''.join(List)

# Function to remove duplicates in a sorted array
def removeDupsSorted(List):
    res_ind = 1
    ip_ind = 1

    # In place removal of duplicate characters
    while ip_ind != len(List):
        if List[ip_ind] != List[ip_ind-1]:
            List[res_ind] = List[ip_ind]
            res_ind += 1
        ip_ind+=1

    # After above step string is efgkorskkorss.
    # Removing extra kkorss after string
    string = toString(List[0:res_ind])

    return string

# Function removes duplicate characters from the string
# This function work in-place and fills null characters
# in the extra space left
def removeDups(string):
    # Convert string to list
    List = toMutable(string)

    # Sort the character list
    List.sort()

    # Remove duplicates from sorted
    return removeDupsSorted(List)

# Driver program to test the above functions
string = "geeksforgeeks"
print(removeDups(string))

# This code is contributed by Bhavya Jain

C#

// C# program to remove duplicates, the order of
// characters is not maintained in this program
using System;

class GFG
{
    /* Method to remove duplicates in a sorted array */
    static String removeDupsSorted(String str)
    {
        int res_ind = 1, ip_ind = 1;

        // Character array for removal of duplicate characters
        char []arr = str.ToCharArray();

        /* In place removal of duplicate characters*/
        while (ip_ind != arr.Length)
        {
            if(arr[ip_ind] != arr[ip_ind-1])
            {
                arr[res_ind] = arr[ip_ind];
                res_ind++;
            }
            ip_ind++;

        }

        str = new String(arr);
        return str.Substring(0,res_ind);
    }

    /* Method removes duplicate characters from the string
    This function work in-place and fills null characters
    in the extra space left */
    static String removeDups(String str)
    {
    // Sort the character array
    char []temp = str.ToCharArray();
    Array.Sort(temp);
    str = String.Join("",temp);

    // Remove duplicates from sorted
    return removeDupsSorted(str);
    }

    // Driver Method
    public static void Main(String[] args)
    {
        String str = "geeksforgeeks";
        Console.WriteLine(removeDups(str));
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

function removeDuplicate(string)
{
   return string.split('')
    .filter(function(item, pos, self)
    {
      return self.indexOf(item) == pos;
    }
   ).join('');
}

var str = "geeksforgeeks";
document.write( " "+removeDuplicate(str));

//This code is contributed by SoumikMondal
</script>

Output

efgkors

Time Complexity: O(n log n) If we use some nlogn sorting algorithm instead of quicksort.
Auxiliary Space: O(1)

METHOD 4 (Use Hashing )

Algorithm:

1: Initialize:
    str  =  "test string" /* input string */
    ip_ind =  0          /* index to  keep track of location of next
                             character in input string */
    res_ind  =  0         /* index to  keep track of location of
                            next character in the resultant string */
    bin_hash[0..255] = {0,0, ….} /* Binary hash to see if character is 
                                        already processed or not */
2: Do following for each character *(str + ip_ind) in input string:
              (a) if bin_hash is not set for *(str + ip_ind) then
                   // if program sees the character *(str + ip_ind) first time
                         (i)  Set bin_hash for *(str + ip_ind)
                         (ii)  Move *(str  + ip_ind) to the resultant string.
                              This is done in-place.
                         (iii) res_ind++
              (b) ip_ind++
  /* String obtained after this step is "the stringing" */
3: Remove extra characters at the end of the resultant string.
  /*  String obtained after this step is "te string" */

Implementation:

Code block

Output

geksfor

Time Complexity: O(n)
Auxiliary Space: O(1)

Important Points:

• Method 2 doesn’t maintain the characters as original strings, but method 4 does.
• It is assumed that the number of possible characters in the input string is 256. NO_OF_CHARS should be changed accordingly.
• calloc() is used instead of malloc() for memory allocations of a counting array (count) to initialize allocated memory to ‘\0’. the malloc() followed by memset() could also be used.
• The above algorithm also works for integer array inputs if the range of the integers in the array is given. An example problem is to find the maximum occurring number in an input array given that the input array contains integers only between 1000 to 1100

Method 5 (Using IndexOf() method) : 
Prerequisite : Java IndexOf() method

C++

// C++ program to create a unique string
#include <bits/stdc++.h>
using namespace std;

// Function to make the string unique
string unique(string s)
{
    string str;
    int len = s.length();

    // loop to traverse the string and
    // check for repeating chars using
    // IndexOf() method in Java
    for(int i = 0; i < len; i++)
    {

        // character at i'th index of s
        char c = s[i];

        // If c is present in str, it returns
        // the index of c, else it returns npos
        auto found = str.find(c);
        if (found == std::string::npos)
        {

            // Adding c to str if npos is returned
            str += c;
        }
    }
    return str;
}

// Driver code
int main()
{

    // Input string with repeating chars
    string s = "geeksforgeeks";

    cout << unique(s) << endl;
}

// This code is contributed by nirajgusain5

Java

// Java program to create a unique string
import java.util.*;

class IndexOf {

    // Function to make the string unique
    public static String unique(String s)
    {
        String str = new String();
        int len = s.length();

        // loop to traverse the string and
        // check for repeating chars using
        // IndexOf() method in Java
        for (int i = 0; i < len; i++)
        {
            // character at i'th index of s
            char c = s.charAt(i);

            // if c is present in str, it returns
            // the index of c, else it returns -1
            if (str.indexOf(c) < 0)
            {
                // adding c to str if -1 is returned
                str += c;
            }
        }

        return str;
    }

    // Driver code
    public static void main(String[] args)
    {
        // Input string with repeating chars
        String s = "geeksforgeeks";

        System.out.println(unique(s));
    }
}

Python3

# Python 3 program to create a unique string

# Function to make the string unique


def unique(s):

    st = ""
    length = len(s)

    # loop to traverse the string and
    # check for repeating chars using
    # IndexOf() method in Java
    for i in range(length):

        # character at i'th index of s
        c = s[i]

        # if c is present in str, it returns
        # the index of c, else it returns - 1
        # print(st.index(c))
        if c not in st:
            # adding c to str if -1 is returned
            st += c

    return st


# Driver code
if __name__ == "__main__":

    # Input string with repeating chars
    s = "geeksforgeeks"

    print(unique(s))

    # This code is contributed by ukasp.

C#

// C# program to create a unique string
using System;

public class IndexOf
{

    // Function to make the string unique
    public static String unique(String s)
    {
        String str = "";
        int len = s.Length;

        // loop to traverse the string and
        // check for repeating chars using
        // IndexOf() method in Java
        for (int i = 0; i < len; i++)
        {
            // character at i'th index of s
            char c = s[i];

            // if c is present in str, it returns
            // the index of c, else it returns -1
            if (str.IndexOf(c) < 0)
            {
                // adding c to str if -1 is returned
                str += c;
            }
        }

        return str;
    }

    // Driver code
    public static void Main(String[] args)
    {
        // Input string with repeating chars
        String s = "geeksforgeeks";

        Console.WriteLine(unique(s));
    }
}

// This code is contributed by Princi Singh

Javascript

<script>

    // JavaScript program to create a unique string

    // Function to make the string unique
    function unique(s)
    {
        let str = "";
        let len = s.length;

        // loop to traverse the string and
        // check for repeating chars using
        // IndexOf() method in Java
        for (let i = 0; i < len; i++)
        {
            // character at i'th index of s
            let c = s[i];

            // if c is present in str, it returns
            // the index of c, else it returns -1
            if (str.indexOf(c) < 0)
            {
                // adding c to str if -1 is returned
                str += c;
            }
        }

        return str;
    }

      // Input string with repeating chars
    let s = "geeksforgeeks";

    document.write(unique(s));

</script>

Output

geksfor

Time Complexity: O(n)
Auxiliary Space: O(n), where n is the size of the given string.

Thanks debjitdbb for suggesting this approach.

Method 6 (Using unordered_map STL method) : 
Prerequisite : unordered_map STL C++ method

C++

// C++ program to create a unique string using unordered_map

/* access time in unordered_map on is O(1) generally if no collisions occur
and therefore it helps us check if an element exists in a string in O(1)
time complexity with constant space. */

#include <bits/stdc++.h>
using namespace std;
char* removeDuplicates(char *s,int n){
  unordered_map<char,int> exists;
  int index = 0;
  for(int i=0;i<n;i++){
    if(exists[s[i]]==0)
    {
      s[index++] = s[i];
      exists[s[i]]++;
    }
  }
  return s;
}

//driver code
int main(){
  char s[] = "geeksforgeeks";
  int n = sizeof(s)/sizeof(s[0]);
  cout<<removeDuplicates(s,n)<<endl;
  return 0;
}

Java

// Java program to create a unique String using unordered_map

/* access time in unordered_map on is O(1) generally if no collisions occur
and therefore it helps us check if an element exists in a String in O(1)
time complexity with constant space. */
import java.util.*;

class GFG{
static char[] removeDuplicates(char []s,int n){
  Map<Character,Integer> exists = new HashMap<>();

  String st = "";
  for(int i = 0; i < n; i++){
    if(!exists.containsKey(s[i]))
    {
      st += s[i];
      exists.put(s[i], 1);
    }
  }
  return st.toCharArray();
}

// driver code
public static void main(String[] args){
  char s[] = "geeksforgeeks".toCharArray();
  int n = s.length;
  System.out.print(removeDuplicates(s,n));
}
}

Python3

# Python program to create a unique string using unordered_map

# access time in unordered_map on is O(1) generally if no collisions occur
# and therefore it helps us check if an element exists in a string in O(1)
# time complexity with constant space.
def removeDuplicates(s, n):
    exists = {}
    index = 0
    ans = ""

    for i in range(0, n):
        if s[i] not in exists or exists[s[i]] == 0:
            s[index] = s[i]
            print(s[index], end='')
            index += 1
            exists[s[i]] = 1

# driver code
s = "geeksforgeeks"
s1 = list(s)
n = len(s1)
removeDuplicates(s1, n)

C#

// C# program to create a unique String using unordered_map

/* access time in unordered_map on is O(1) generally if no collisions occur
and therefore it helps us check if an element exists in a String in O(1)
time complexity with constant space. */
using System;
using System.Collections.Generic;

public class GFG{
static char[] removeDuplicates(char []s,int n){
  Dictionary<char,int> exists = new Dictionary<char, int>();

  String st = "";
  for(int i = 0; i < n; i++){
    if(!exists.ContainsKey(s[i]))
    {
      st += s[i];
      exists.Add(s[i], 1);
    }
  }
  return st.ToCharArray();
}

// driver code
public static void Main(String[] args){
  char []s = "geeksforgeeks".ToCharArray();
  int n = s.Length;
  Console.Write(removeDuplicates(s,n));
}
}

Javascript

<script>
// javascript program to create a unique String using unordered_map

/* access time in unordered_map on is O(1) generally if no collisions occur
and therefore it helps us check if an element exists in a String in O(1)
time complexity with constant space. */
     function removeDuplicates( s , n) {
        var exists = new Map();

        var st = "";
        for (var i = 0; i < n; i++) {
            if (!exists.has(s[i])) {
                st += s[i];
                exists.set(s[i], 1);
            }
        }
        return st;
    }

    // driver code

        var s = "geeksforgeeks";
        var n = s.length;
        document.write(removeDuplicates(s, n));

</script>

Output

geksfor

Time Complexity: O(n)
Auxiliary Space: O(n)
Thanks, Allen James Vinoy for suggesting this approach.

METHOD 7 (Simple) 

C++

#include <iostream>
using namespace std;

int main()
{
    string s = "abcdabd";
    string temp = "";
    temp += s.at(0);

    for (int i = 1; i < s.length(); i++) {
        if (!(temp.find(s.at(i)) < temp.length()))
            temp = temp + s.at(i);
    }
    cout << temp;
    return 0;
}

Java

import java.io.*;

class GFG {
    public static void main(String[] args)
    {
        String s = "abcdabd";
        String temp = "" + s.charAt(0);

        for (int i = 1; i < s.length(); i++) {
            if (!temp.contains(String.valueOf(s.charAt(i))))
                temp = temp + s.charAt(i);
        }
        System.out.println(temp);
    }
}

Python3

# Python code for the above Approach
s = "abcdabd"
temp = "" + s[0]

for i in range(1,len(s)):
  if (s[i] not in temp):
    temp = temp + s[i]

# Printing the Result
print(temp)

C#

using System;

public class GFG{

  static public void Main (){
    string s = "abcdabd";
    string temp = "";
    temp += s[0];

    for (int i = 1; i < s.Length; i++) {
      if (temp.IndexOf(s[i]) == -1)
        temp = temp + s[i];
    }
    System.Console.WriteLine(temp);
  }
}

// This code is contributed by akashish__

Javascript

<script>

// JavaScript code for the above Approach
let s = "abcdabd"
let temp = "" + s[0]

for(let i=1;i<s.length;i++){
    if (temp.indexOf(s[i]) == -1)
        temp = temp + s[i]
}

// Printing the Result
document.write(temp,"</br>")

</script>

Output

abcd

Time Complexity: O(n)
Auxiliary Space: O(n)