Java Program To Remove Duplicates From A Given String
Last Updated :
23 Jul, 2025
Write a java program for a given 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.
Note: The order of remaining characters in the output should be the same as in the original string.
Example:
Input: Str = geeksforgeeks
Output: geksfor
Explanation: After removing duplicate characters such as e, k, g, s, we have string as “geksfor”.
Input: Str = HappyNewYear
Output: HapyNewYr
Explanation: After removing duplicate characters such as p, e, a, we have string as “HapyNewYr”.
Naive Approach:
Iterate through the string and for each character check if that particular character has occurred before it in the string. If not, add the character to the result, otherwise the character is not added to result.
Below is the implementation of above approach:
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
Output:
geksfor
Time Complexity : O(n * n)
Auxiliary Space : O(1) , Keeps order of elements the same as input.
Remove duplicates from a given string using Hashing
Iterating through the given string and use a map to efficiently track of encountered characters. If a character is encountered for the first time, it’s added to the result string, Otherwise, it’s skipped. This ensures the output string contains only unique characters in the same order as the input string.
Below is the implementation of above approach:
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));
}
}
Output:
geksfor
Time Complexity: O(n)
Space complexity: O(n)
Please refer complete article on Remove duplicates from a given string for more details!
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