Remove duplicates from a string in O(1) extra space

12 Jul, 2022
Given a string str of lowercase characters, the task is to remove duplicates and return a resultant string without modifying the order of characters in the original string.
Examples:  
Input: str = "geeksforgeeks"
Output: geksfor

Input: str = "characters"
Output: chartes
Approach: The idea is to use bits of a counter variable to mark the presence of a character in the string. To mark the presence of ‘a’ set 0th bit as 1, for ‘b’ set 1st bit as 1 and so on. If the corresponding bit of character present in the original string is set to 0, it means it is the first occurrence of that character, hence set its corresponding bit as 1 and keep on including the current character in the resultant string.
Consider the string str = “geeksforgeeks” 
character: ‘g’ x = 6(ascii of g – 97) 6th bit in counter is unset resulting first occurrence of character ‘g’. str[0] = ‘g’ counter = 00000000000000000000000001000000 // mark 6th bit as visited length = 1
character: ‘e’ x = 4(ascii of e – 97) 4th bit in counter is unset resulting in first occurrence of character ‘e’. str[1] = ‘e’ counter = 00000000000000000000000001010000 //mark 4th bit as visited length = 2
character: ‘e’ x = 4(ascii of e – 97) 4th bit in counter is set resulting in duplicate character. Ignore this character. Move for next character. counter = 00000000000000000000000001010000 //same as previous length = 2
character: ‘k’ x = 10(ascii of k – 97) 10th bit in counter is unset resulting in first occurrence of character ‘k’. str[2] = ‘k’ counter = 00000000000000000000010001010000 //mark 10th bit as visited length = 3
Similarly, do the same for all characters.Resultant string : geksfor(string of length 7 starting from index 0)  
Algorithm: 
Initialize a counter variable (keeps track of the characters visited in string), it is a 32 bit Integer represented as(00000000000000000000000000000000) initially.Consider ‘a’ as 0th bit of counter, ‘b’ as 1st bit of counter, ‘c’ as 2nd bit of counter and so on.Traverse through each character of input string.Get the character’s value, where character’s value(x) = Ascii of character – 97. This will make sure for value of ‘a’ as 0, value of ‘b’ as 1 and so on.Check xth bit of counter.If Xth bit of counter is 0 which means the current character has appeared for the first time, keep the current character at the index “length” of string .Mark the current character visited by setting xth bit of counter.Increment length.Return Substring of size “length” from index 0.
Initialize a counter variable (keeps track of the characters visited in string), it is a 32 bit Integer represented as(00000000000000000000000000000000) initially.
Consider ‘a’ as 0th bit of counter, ‘b’ as 1st bit of counter, ‘c’ as 2nd bit of counter and so on.
Traverse through each character of input string.
Get the character’s value, where character’s value(x) = Ascii of character – 97. This will make sure for value of ‘a’ as 0, value of ‘b’ as 1 and so on.
Check xth bit of counter.
If Xth bit of counter is 0 which means the current character has appeared for the first time, keep the current character at the index “length” of string .
Mark the current character visited by setting xth bit of counter.
Increment length.
Return Substring of size “length” from index 0.
Below is the implementation of above approach: 
C++
Java
Python3
C#
PHP
Javascript
// C++ implementation of above approach#include <bits/stdc++.h>#include <string>using namespace std; // Function to remove duplicatesstring removeDuplicatesFromString(string str){     // keeps track of visited characters    int counter = 0;     int i = 0;    int size = str.size();     // gets character value    int x;     // keeps track of length of resultant string    int length = 0;     while (i < size) {        x = str[i] - 97;         // check if Xth bit of counter is unset        if ((counter & (1 << x)) == 0) {             str[length] = 'a' + x;             // mark current character as visited            counter = counter | (1 << x);             length++;        }        i++;    }     return str.substr(0, length);} // Driver codeint main(){    string str = "geeksforgeeks";    cout << removeDuplicatesFromString(str);    return 0;}
// Java implementation of above approachimport java.util.Arrays; class GFG {     // Function to remove duplicates    static char[] removeDuplicatesFromString(String string)    {         // keeps track of visited characters        int counter = 0;        char[] str = string.toCharArray();        int i = 0;        int size = str.length;         // gets character value        int x;         // keeps track of length of resultant String        int length = 0;         while (i < size) {            x = str[i] - 97;             // check if Xth bit of counter is unset            if ((counter & (1 << x)) == 0) {                 str[length] = (char)('a' + x);                 // mark current character as visited                counter = counter | (1 << x);                 length++;            }            i++;        }         return Arrays.copyOfRange(str, 0, length);    }     // Driver code    public static void main(String[] args)    {        String str = "geeksforgeeks";        System.out.println(removeDuplicatesFromString(str));    }} // This code is contributed by Mithun Kumar
# Python3 implementation of above approach # Function to remove duplicatesdef removeDuplicatesFromString(str2):     # keeps track of visited characters    counter = 0;     i = 0;    size = len(str2);    str1 = list(str2);     # gets character value    x = 0;     # keeps track of length of resultant string    length = 0;     while (i < size):        x = ord(str1[i]) - 97;         # check if Xth bit of counter is unset        if ((counter & (1 << x)) == 0):            str1[length] = chr(97 + x);             # mark current character as visited            counter = counter | (1 << x);             length += 1;        i += 1;             str2=''.join(str1);    return str2[0:length]; # Driver codestr1 = "geeksforgeeks";print(removeDuplicatesFromString(str1)); # This code is contributed by mits
// C# implementation of above approachusing System; class GFG {     // Function to remove duplicates    static string removeDuplicatesFromString(string string1)    {         // keeps track of visited characters        int counter = 0;        char[] str = string1.ToCharArray();        int i = 0;        int size = str.Length;         // gets character value        int x;         // keeps track of length of resultant String        int length = 0;         while (i < size) {            x = str[i] - 97;             // check if Xth bit of counter is unset            if ((counter & (1 << x)) == 0) {                 str[length] = (char)('a' + x);                 // mark current character as visited                counter = counter | (1 << x);                 length++;            }            i++;        }         return (new string(str)).Substring(0, length);    }     // Driver code    static void Main()    {        string str = "geeksforgeeks";        Console.WriteLine(removeDuplicatesFromString(str));    }} // This code is contributed by mits
<?php// PHP implementation of above approach // Function to remove duplicatesfunction removeDuplicatesFromString($str){     // keeps track of visited characters    $counter = 0;     $i = 0;    $size = strlen($str);     // gets character value    $x = 0;     // keeps track of length of resultant string    $length = 0;     while ($i < $size)    {        $x = ord($str[$i]) - 97;         // check if Xth bit of counter is unset        if (($counter & (1 << $x)) == 0)        {            $str[$length] = chr(97 + $x);             // mark current character as visited            $counter = $counter | (1 << $x);             $length++;        }        $i++;    }     return substr($str, 0, $length);} // Driver code$str = "geeksforgeeks";echo removeDuplicatesFromString($str); // This code is contributed by mits?>
<script>// Javascript implementation of above approach // Function to remove duplicatesfunction removeDuplicatesFromString(string){    // keeps track of visited characters        let counter = 0;        let str = string.split("");        let i = 0;        let size = str.length;          // gets character value        let x;          // keeps track of length of resultant String        let length = 0;          while (i < size) {            x = str[i].charCodeAt(0) - 97;              // check if Xth bit of counter is unset            if ((counter & (1 << x)) == 0) {                  str[length] = String.fromCharCode('a'.charCodeAt(0) + x);                  // mark current character as visited                counter = counter | (1 << x);                  length++;            }            i++;        }          return str.join("").slice(0,length);} // Driver codelet  str = "geeksforgeeks";document.write(removeDuplicatesFromString(str));  // This code is contributed by patel2127</script>
geksfor
Time Complexity: O(n) Space Complexity: O(n) -> As it uses char[] array of string to store characters of string (i.e. dependent upon length of input string)
Another Approach: This approach keeps track of visited characters from given input string through an integer array of size 256 (All possible characters).The idea is as follows: 
Create an integer array of size 256 in order to keep track of all possible characters.Iterate over the input string, and for each character :Lookup into the array with the ASCII value of character as index: If value at index is 0, then copy the character into original input array and increase the endIndex also update the value at index as -1.Else skip the character.
Create an integer array of size 256 in order to keep track of all possible characters.
Iterate over the input string, and for each character :
Lookup into the array with the ASCII value of character as index: If value at index is 0, then copy the character into original input array and increase the endIndex also update the value at index as -1.Else skip the character.
If value at index is 0, then copy the character into original input array and increase the endIndex also update the value at index as -1.
Else skip the character.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std; // Method to remove duplicatesstring removeDuplicatesFromString(string str){         // Table to keep track of visited characters    vector<int> table(256, 0);    vector<char> chars;         for(auto i : str)        chars.push_back(i);         // To keep track of end index of    // resultant string    int endIndex = 0;         for(int i = 0; i < chars.size(); i++)    {        if (table[chars[i]] == 0)        {            table[chars[i]] = -1;            chars[endIndex++] = chars[i];        }    }         string ans = "";         for(int i = 0; i < endIndex; i++)        ans += chars[i];         return ans;} // Driver codeint main(){    string str = "geeksforgeeks";         cout << (removeDuplicatesFromString(str))         << endl;} // This code is contributed by Mohit kumar 29
//Java implementation of above approachimport java.util.Arrays; class GFG {     // Method to remove duplicates    static char[] removeDuplicatesFromString(String string)    {        //table to keep track of visited characters        int[] table = new int[256];        char[] chars = string.toCharArray();         //to keep track of end index of resultant string        int endIndex = 0;             for(int i = 0; i < chars.length; i++)        {            if(table[chars[i]] == 0)            {                table[chars[i]] = -1;                chars[endIndex++] = chars[i];            }        }             return Arrays.copyOfRange(chars, 0, endIndex);    }     // Driver code    public static void main(String[] args)    {        String str = "geeksforgeeks";        System.out.println(removeDuplicatesFromString(str));    }} // This code is contributed by Sonu Singh
# Python3 implementation of above approach # Method to remove duplicatesdef removeDuplicatesFromString(string):         # Table to keep track of visited    # characters    table = [0 for i in range(256)]         # To keep track of end index    # of resultant string    endIndex = 0    string = list(string)         for i in range(len(string)):        if (table[ord(string[i])] == 0):            table[ord(string[i])] = -1            string[endIndex] = string[i]            endIndex += 1                 ans = ""    for i in range(endIndex):      ans += string[i]           return ans # Driver codeif __name__ == '__main__':         temp = "geeksforgeeks"         print(removeDuplicatesFromString(temp))     # This code is contributed by Kuldeep Singh
// C# implementation of above approachusing System; class GFG{     // Method to remove duplicates    static char[] removeDuplicatesFromString(String str)    {        // table to keep track of visited characters        int[] table = new int[256];        char[] chars = str.ToCharArray();         // to keep track of end index        // of resultant string        int endIndex = 0;             for(int i = 0; i < chars.Length; i++)        {            if(table[chars[i]] == 0)            {                table[chars[i]] = -1;                chars[endIndex++] = chars[i];            }        }        char []newStr = new char[endIndex];        Array.Copy(chars, newStr, endIndex);        return newStr;    }     // Driver code    public static void Main(String[] args)    {        String str = "geeksforgeeks";        Console.WriteLine(removeDuplicatesFromString(str));    }} // This code is contributed by 29AjayKumar
<script> // JavaScript implementation of above approach     // Method to remove duplicates    function removeDuplicatesFromString(string)    {        // table to keep track of visited characters        let table = new Array(256);        for(let i=0;i<table.length;i++)            table[i]=0;        let chars = string.split("");          // to keep track of end index of resultant string        let endIndex = 0;              for(let i = 0; i < chars.length; i++)        {            if(table[chars[i].charCodeAt(0)] == 0)            {                                 table[chars[i].charCodeAt(0)] = -1;                chars[endIndex++] = chars[i];                             }        }               let ans="";         for(let i=0;i<endIndex;i++)            ans += chars[i]        return ans;    }     // Driver code    let str = "geeksforgeeks";    document.write(removeDuplicatesFromString(str));      // This code is contributed by unknown2108 </script>
geksfor
Time Complexity: O(n) Space Complexity: O(n) -> As it uses char[] array of string to store characters of string (i.e. dependent upon length of input string)This approach is contributed by Sonu Singh.  
Another Approach:
In cases where adding space is especially expensive, we can utilize the following approach:
1. Iterate over the string letter by letter.
2. Find the first occurrence of the letter at the ith position of the string. This can be done using in-built methods such as find in C++ STL, indexOf in Java, index in Python, and indexOf in JavaScript.
3. If the first occurrence is not equal to i, then delete the character from the string. This can be done using in-built methods such as erase in C++ STL, and deleteCharAt in Java, or using built-in methods to construct a substring.
C++
Python3
Javascript
// C++ program to create a string of only unique characters#include <bits/stdc++.h>using namespace std;  // Function to make the string uniquestring removeDuplicates(string s){     // loop to traverse the string and    //and check if the first occurrence of    //each letter matches the current index    for(int i = 0; i < s.length(); i++)    {        // if the first occurrence of s[i]        //is not i, then remove it from s        if (s.find(s[i]) != i)        {            s.erase(i, 1);            i--;        }    }    return s;}  // Driver codeint main(){          // Input string with repeating chars    string s = "geeksforgeeks";      cout << removeDuplicates(s) << endl;} //This code is contributed by phasing17
# Python3 program to create a string of only unique# characters # Function to make the string uniquedef removeDuplicates(s):     n = len(s)         # loop to traverse the string and    # and check if the first occurrence of    # each letter matches the current index    i = 0    while i < n:         # if the first occurrence of s[i]        # is not i, then remove it from s        if (s.index(s[i]) != i):            s = s[:i] + s[i + 1:]            i -= 1            n -= 1        else:            i += 1     return s # Driver code # Input string with repeating charss = "geeksforgeeks" # Function Callprint(removeDuplicates(s)) # This code is contributed by phasing17
// JavaScript program to create a string of only unique// characters // Function to make the string uniquefunction removeDuplicates(s){       // loop to traverse the string and    // and check if the first occurrence of    // each letter matches the current index    for (var i = 0; i < s.length; i++) {               // if the first occurrence of s[i]        // is not i, then remove it from s        if (s.indexOf(s[i]) != i) {            s = s.slice(0, i) + s.slice(i + 1, s.length);            i--;        }    }    return s;} // Driver code // Input string with repeating charslet s = "geeksforgeeks"; // Function Callconsole.log(removeDuplicates(s)); // This code is contributed by phasing17
geksfor
Time Complexity: O(N2) 
Auxiliary Space: O(1)
Rajput-Ji
Mithun Kumar
SonuSinghRajput1
29AjayKumar
durgesh khandal
jordon50797
mohit kumar 29
patel2127
unknown2108
phasing17
satyajitmishra
sweetyty
String Duplicates
Bit Magic
Strings
Strings
Bit Magic
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