Remove duplicates from a string in O(1) extra space
Last Updated :
01 Sep, 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.
Below is the implementation of above approach:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
#include <string>
using namespace std;
// Function to remove duplicates
string 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 code
int main()
{
string str = "geeksforgeeks";
cout << removeDuplicatesFromString(str);
return 0;
}
Java
// Java implementation of above approach
import 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
# Python3 implementation of above approach
# Function to remove duplicates
def 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 code
str1 = "geeksforgeeks";
print(removeDuplicatesFromString(str1));
# This code is contributed by mits
C#
// C# implementation of above approach
using 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
// PHP implementation of above approach
// Function to remove duplicates
function 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
?>
JavaScript
<script>
// Javascript implementation of above approach
// Function to remove duplicates
function 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 code
let str = "geeksforgeeks";
document.write(removeDuplicatesFromString(str));
// This code is contributed by patel2127
</script>
Complexity Analysis:
- 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.
Below is the implementation of the above approach:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Method to remove duplicates
string 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 code
int main()
{
string str = "geeksforgeeks";
cout << (removeDuplicatesFromString(str))
<< endl;
}
// This code is contributed by Mohit kumar 29
Java
//Java implementation of above approach
import 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
# Python3 implementation of above approach
# Method to remove duplicates
def 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 code
if __name__ == '__main__':
temp = "geeksforgeeks"
print(removeDuplicatesFromString(temp))
# This code is contributed by Kuldeep Singh
C#
// C# implementation of above approach
using 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
JavaScript
<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>
Complexity AnaysAnalysisis:
- 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:
- Iterate over the string letter by letter.
- 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.
- 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.
Implementation:
C++
// C++ program to create a string of only unique characters
#include <bits/stdc++.h>
using namespace std;
// Function to make the string unique
string 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 code
int main()
{
// Input string with repeating chars
string s = "geeksforgeeks";
cout << removeDuplicates(s) << endl;
}
//This code is contributed by phasing17
Java
// Java program to create a string of only unique characters
import java.util.*;
class GFG {
// Function to make the string unique
static String 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.indexOf(s.charAt(i)) != i) {
StringBuilder sb = new StringBuilder(s);
sb.deleteCharAt(i);
s = sb.toString();
i--;
}
}
return s;
}
// Driver code
public static void main(String[] args)
{
// Input string with repeating chars
String s = "geeksforgeeks";
System.out.println(removeDuplicates(s));
}
}
// This code is contributed by phasing17
Python3
# Python3 program to create a string of only unique
# characters
# Function to make the string unique
def 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 chars
s = "geeksforgeeks"
# Function Call
print(removeDuplicates(s))
# This code is contributed by phasing17
C#
// C# program to create a string of only unique characters
using System;
using System.Collections.Generic;
class GFG
{
// Function to make the string unique
static string 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.IndexOf(s[i]) != i) {
s = s.Remove(i, 1);
i--;
}
}
return s;
}
// Driver code
public static void Main(string[] args)
{
// Input string with repeating chars
string s = "geeksforgeeks";
Console.WriteLine(removeDuplicates(s));
}
}
// This code is contributed by phasing17
JavaScript
// JavaScript program to create a string of only unique
// characters
// Function to make the string unique
function 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 chars
let s = "geeksforgeeks";
// Function Call
console.log(removeDuplicates(s));
// This code is contributed by phasing17
Complexity Analysis:
- Time Complexity: O(N2)
- Auxiliary Space: O(1)
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