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Print characters having odd frequencies in order of occurrence

  • Last Updated : 12 May, 2021

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 
 

CharacterFrequency
‘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 characters 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++




// 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;
}

Java




// 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.

Python3




# 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

C#




// 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

Javascript




<script>
// javascript implementation of the approach
 
let SIZE = 26;
 
// Function to print the odd frequency characters
// in the order of their occurrence
function printChar(str, n)
{
    // To store the frequency of each of
    // the character of the string
    let freq = [];
 
    // Initialize all elements of freq[] to 0
    for(let i = 0; i < SIZE; i++){
        freq.push(0);
    }
 
    // Update the frequency of each character
    for (let i = 0; i < n; i++)
        freq[str.charCodeAt(i) - 97]++;
 
    // Traverse str character by character
    for (let i = 0; i < n; i++) {
 
        // If frequency of current character is odd
        if (freq[str.charCodeAt(i) - 97] % 2 == 1) {
            document.write(str[i]);
        }
    }
}
 
let str = "geeksforgeeks";
let n = str.length;
printChar(str, n);
 
// This code is contributed by rohitsingh07052.
</script>

Output: 



for

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

Method #2: Using built-in python functions.

Approach:

We will scan the string and count the occurrence of all characters  using built in Counter() function after that we  traverse the counter list and check if the occurrences are odd or not  if there is any even frequency character then we immediately print No.

Note: This method is applicable for all type of characters

Below is the implementation of the above approach:

Python3




# importing Counter function
from collections import Counter
 
# function to check if all
# elements occur odd times
def checkString(s):
   
    # Counting the frequency of all
    # character using Counter function
    frequency = Counter(s)
     
    # Traversing frequency
    for i in frequency:
       
        # Checking if any element
        # has even count
        if (frequency[i] % 2 == 0):
            return False
    return True
 
 
# Driver code
s = "gggfffaaa"
if(checkString(s)):
    print("Yes")
else:
    print("No")
     
# This code is contributed by vikkycirus

Output:

Yes

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