Print characters having odd frequencies in order of occurrence

Given a string str containing only lowercase characters. The task is to print the characters having an odd frequency in the order of their occurrence.

Note: Repeated elements with odd frequency are printed as many times they occur in order of their occurrences.

Examples:



Input: str = “geeksforgeeks”
Output: for

Character Frequency
‘g’ 2
‘e’ 4
‘k’ 2
‘s’ 2
‘f’ 1
‘o’ 1
‘r’ 1

‘f’, ‘o’ and ‘r’ are the only characters with odd frequencies.

Input: str = “elephant”
Output: lphant

Approach: Create a frequency array to store the frequency of each of the character of the given string str. Traverse the string str again and check whether the frequency of that character is odd. If yes, then print the character.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define SIZE 26
  
// Function to print the odd frequency characters
// in the order of their occurrence
void printChar(string str, int n)
{
  
    // To store the frequency of each of
    // the character of the string
    int freq[SIZE];
  
    // Initialize all elements of freq[] to 0
    memset(freq, 0, sizeof(freq));
  
    // Update the frequency of each character
    for (int i = 0; i < n; i++)
        freq[str[i] - 'a']++;
  
    // Traverse str character by character
    for (int i = 0; i < n; i++) {
  
        // If frequency of current character is odd
        if (freq[str[i] - 'a'] % 2 == 1) {
            cout << str[i];
        }
    }
}
  
// Driver code
int main()
{
    string str = "geeksforgeeks";
    int n = str.length();
    printChar(str, n);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
class GFG {
    // Function to print the odd frequency characters
    // in the order of their occurrence
    public static void printChar(String str, int n)
    {
  
        // To store the frequency of each of
        // the character of the string
        int[] freq = new int[26];
  
        // Update the frequency of each character
        for (int i = 0; i < n; i++)
            freq[str.charAt(i) - 'a']++;
  
        // Traverse str character by character
        for (int i = 0; i < n; i++) {
  
            // If frequency of current character is odd
            if (freq[str.charAt(i) - 'a'] % 2 == 1) {
                System.out.print(str.charAt(i));
            }
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
        String str = "geeksforgeeks";
        int n = str.length();
        printChar(str, n);
    }
}
  
// This code is contributed by Naman_Garg.

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
import sys
import math
  
# Function to print the odd frequency characters 
# in the order of their occurrence 
def printChar(str_, n):
  
    # To store the frequency of each of 
    # the character of the string and
    # Initialize all elements of freq[] to 0 
    freq = [0] * 26
  
    # Update the frequency of each character 
    for i in range(n):
        freq[ord(str_[i]) - ord('a')] += 1
      
    # Traverse str character by character 
    for i in range(n):
  
        # If frequency of current character is odd 
        if (freq[ord(str_[i]) - 
                 ord('a')]) % 2 == 1:
            print("{}".format(str_[i]), end = "")
  
# Driver code
if __name__=='__main__':
    str_ = "geeksforgeeks"
    n = len(str_)
    printChar(str_, n)
  
# This code is contributed by Vikash Kumar 37

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG {
    // Function to print the odd frequency characters
    // in the order of their occurrence
    public static void printChar(String str, int n)
    {
  
        // To store the frequency of each of
        // the character of the string
        int[] freq = new int[26];
  
        // Update the frequency of each character
        for (int i = 0; i < n; i++)
            freq[str[i] - 'a']++;
  
        // Traverse str character by character
        for (int i = 0; i < n; i++) {
  
            // If frequency of current character is odd
            if (freq[str[i] - 'a'] % 2 == 1) {
                Console.Write(str[i]);
            }
        }
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        String str = "geeksforgeeks";
        int n = str.Length;
        printChar(str, n);
    }
}
  
// This code has been contributed by 29AjayKumar

chevron_right



Output:

for

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



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.