# Print characters having even frequencies in order of occurrence

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

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

Examples:

Input: str = “geeksforgeeks”
Output: geeksgeeks

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

‘g’, ‘e’, ‘k’ and ‘s’ are the only characters with even frequencies.

Input: str = “aeroplane”
Output: aeae

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

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 even. 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 even 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 even ` `        ``if` `(freq[str[i] - ``'a'``] % 2 == 0) { ` `            ``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 ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` `static` `int` `SIZE = ``26``; ` ` `  `// Function to print the even frequency characters ` `// in the order of their occurrence ` `static` `void` `printChar(String str, ``int` `n) ` `{ ` ` `  `    ``// To store the frequency of each of ` `    ``// the character of the string ` `    ``int` `[]freq = ``new` `int``[SIZE]; ` ` `  `    ``// 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 even ` `        ``if` `(freq[str.charAt(i) - ``'a'``] % ``2` `== ``0``) ` `        ``{ ` `            ``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 29AjayKumar `

## Python3

 `# Python3 implementation of the approach ` `SIZE ``=` `26` ` `  `# Function to print the even frequency characters ` `# in the order of their occurrence ` `def` `printChar(string, n): ` ` `  `    ``# To store the frequency of each of ` `    ``# the character of the stringing ` `    ``# Initialize all elements of freq[] to 0 ` `    ``freq ``=` `[``0``] ``*` `SIZE ` ` `  `    ``# Update the frequency of each character ` `    ``for` `i ``in` `range``(``0``, n): ` `        ``freq[``ord``(string[i]) ``-` `ord``(``'a'``)] ``+``=` `1` ` `  `    ``# Traverse string character by character ` `    ``for` `i ``in` `range``(``0``, n):  ` ` `  `        ``# If frequency of current character is even ` `        ``if` `(freq[``ord``(string[i]) ``-`  `                 ``ord``(``'a'``)] ``%` `2` `=``=` `0``): ` `            ``print``(string[i], end ``=` `"") ` `         `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``string ``=` `"geeksforgeeks"` `    ``n ``=` `len``(string) ` `    ``printChar(string, n) ` ` `  `# This code is contributed by Ashutosh450 `

## C#

 `// C# implementation of the approach ` `using` `System; ` `using` `System.Collections.Generic; ` `     `  `class` `GFG  ` `{ ` `static` `int` `SIZE = 26; ` ` `  `// Function to print the even frequency characters ` `// in the order of their occurrence ` `static` `void` `printChar(String str, ``int` `n) ` `{ ` ` `  `    ``// To store the frequency of each of ` `    ``// the character of the string ` `    ``int` `[]freq = ``new` `int``[SIZE]; ` ` `  `    ``// 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 even ` `        ``if` `(freq[str[i] - ``'a'``] % 2 == 0) ` `        ``{ ` `            ``Console.Write(str[i]); ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``String str = ``"geeksforgeeks"``; ` `    ``int` `n = str.Length; ` `    ``printChar(str, n); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

Output:

```geeksgeeks
```

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

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