# 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

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‘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 ``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``;` `        ``// 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

 ``

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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