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

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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

geeksgeeks

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



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