# Remove duplicates from string keeping the order according to last occurrences

Given a string, remove duplicate characters from the string, retaining the last occurrence of the duplicate characters. Assume the characters are case-sensitive.

Examples:

Input : geeksforgeeks
Output : forgeks
Explanation : Please note that we keep only last occurrences of repeating characters in same order as they appear in input. If we see result from right side, we can notice that we keep last ‘s’, then last ‘k’ , and so on.

Input : hi this is sample test
Output : hiampl est
Explanation : Here, the output contains last occurrence of every character, even ” “(spaces), and removing the duplicates. Like in this example, there are 4 spaces count, so we have only the last occurrence of space in it removing the others. And there is only last occurrence of each character without repetition.

Input : Abcda
Output : Abcda

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Solution : Traverse the given string from left to right. For every character check if it appears on right side also. If it does, then do not include in output, otherwise include it.

 // C++ program to remove duplicate character // from character array and print in sorted // order #include using namespace std;    string removeDuplicates(string str) {     // Used as index in the modified string     int n = str.length();        // Traverse through all characters     string res = "";     for (int i = 0; i < n; i++) {            // Check if str[i] is present before it         int j;         for (j = i+1; j < n; j++)             if (str[i] == str[j])                 break;            // If not present, then add it to         // result.         if (j == n)             res = res + str[i];     }     return res; }    // Driver code int main() {     string str = "geeksforgeeks";     cout << removeDuplicates(str);     return 0; }

 // Java program to remove duplicate character  // from character array and print in sorted  // order     import java.util.*;     class GFG{     static String removeDuplicates(String str)  {      // Used as index in the modified String      int n = str.length();         // Traverse through all characters      String res = "";      for (int i = 0; i < n; i++)      {          // Check if str[i] is present before it          int j;          for (j = i + 1; j < n; j++)              if (str.charAt(i) == str.charAt(j))                  break;             // If not present, then add it to          // result.          if (j == n)              res = res + str.charAt(i);      }      return res;  }     // Driver code  public static void main(String[] args)  {      String str = "geeksforgeeks";      System.out.print(removeDuplicates(str));  }  }     // This code is contributed by Rajput-Ji

 # Python3 program to remove duplicate character # from character array and prin sorted # order    def removeDuplicates(str):             # Used as index in the modified string     n = len(str)        # Traverse through all characters     res = ""     for i in range(n):            # Check if str[i] is present before it         j = i + 1         while j < n:             if (str[i] == str[j]):                 break             j += 1            # If not present, then add it to         # result.         if (j == n):             res = res + str[i]     return res    # Driver code if __name__ == '__main__':     str = "geeksforgeeks"     print(removeDuplicates(str))    # This code is contributed by mohit kumar 29

 // C# program to remove duplicate character // from character array and print in sorted // order using System;    class GFG{    static String removeDuplicates(String str) {            // Used as index in the modified String     int n = str.Length;        // Traverse through all characters     String res = "";     for(int i = 0; i < n; i++)      {           // Check if str[i] is present before it        int j;        for(j = i + 1; j < n; j++)           if (str[i] == str[j])               break;           // If not present, then add it to        // result.        if (j == n)            res = res + str[i];     }            return res; }    // Driver code public static void Main(String[] args) {     String str = "geeksforgeeks";     Console.Write(removeDuplicates(str)); } }    // This code is contributed by sapnasingh4991

Output:
forgeks

Time Complexity: O(n*n)

Efficient Solution : The idea is to use hashing.
1) Initialize an empty hash table and res = “”
2) Traverse input string from right to left. If the current character is not present in the hash table, append it to res and insert it in the hash table. Else ignore it.

 // C++ program to remove duplicate character // from character array and print in sorted // order #include using namespace std;    string removeDuplicates(string str) {     // Used as index in the modified string     int n = str.length();        // Create an empty hash table     unordered_set s;        // Traverse through all characters from     // right to left     string res = "";     for (int i = n-1; i >= 0; i--) {           // If current character is not in         if (s.find(str[i]) == s.end())        {           res = res + str[i];           s.insert(str[i]);        }     }        // Reverse the result string     reverse(res.begin(), res.end());        return res; }    // Driver code int main() {     string str = "geeksforgeeks";     cout << removeDuplicates(str);     return 0; }

 // Java program to remove duplicate character // from character array and print in sorted // order import java.util.*;    class GFG{    static String removeDuplicates(String str) {            // Used as index in the modified String     int n = str.length();        // Create an empty hash table     HashSet s = new HashSet();        // Traverse through all characters from     // right to left     String res = "";     for(int i = n - 1; i >= 0; i--)     {                   // If current character is not in         if (!s.contains(str.charAt(i)))        {            res = res + str.charAt(i);            s.add(str.charAt(i));        }     }            // Reverse the result String     res = reverse(res);     return res; }    static String reverse(String input) {            char[] a = input.toCharArray();     int l, r = a.length - 1;            for(l = 0; l < r; l++, r--)     {        char temp = a[l];        a[l] = a[r];        a[r] = temp;     }     return String.valueOf(a); }    // Driver code public static void main(String[] args) {     String str = "geeksforgeeks";     System.out.print(removeDuplicates(str)); } }    // This code is contributed by sapnasingh4991

 # Python3 program to remove duplicate character # from character array and prin sorted order def removeDuplicates(str):            # Used as index in the modified string     n = len(str)            # Create an empty hash table     s = set()            # Traverse through all characters from     # right to left     res = ""     for i in range(n - 1, -1, -1):                    # If current character is not in          if (str[i] not in s):             res = res + str[i]             s.add(str[i])            # Reverse the result string     res = res[::-1]     return res    # Driver code str = "geeksforgeeks"    print(removeDuplicates(str))    # This code is contributed by ShubhamCoder

 // C# program to remove duplicate character // from character array and print in sorted // order using System; using System.Collections.Generic;    class GFG{    static String removeDuplicates(String str) {            // Used as index in the modified String     int n = str.Length;        // Create an empty hash table     HashSet s = new HashSet();        // Traverse through all characters      // from right to left     String res = "";     for(int i = n - 1; i >= 0; i--)     {                    // If current character is not in          if (!s.Contains(str[i]))         {             res = res + str[i];             s.Add(str[i]);         }     }            // Reverse the result String     res = reverse(res);     return res; }    static String reverse(String input) {     char[] a = input.ToCharArray();     int l, r = a.Length - 1;            for(l = 0; l < r; l++, r--)     {         char temp = a[l];         a[l] = a[r];         a[r] = temp;     }     return String.Join("", a); }    // Driver code public static void Main(String[] args) {     String str = "geeksforgeeks";     Console.Write(removeDuplicates(str)); } }    // This code is contributed by Rajput-Ji

Output:
forgeks

Time Complexity: O(n)

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