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 = “neveropen”
Output: neveropenneveropen
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++
// 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 occurrencevoid 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 codeint main(){ string str = "neveropen"; int n = str.length(); printChar(str, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {static int SIZE = 26;// Function to print the even frequency characters// in the order of their occurrencestatic 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 codepublic static void main(String[] args){ String str = "neveropen"; int n = str.length(); printChar(str, n);}} // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approachSIZE = 26# Function to print the even frequency characters# in the order of their occurrencedef 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 codeif __name__ == '__main__': string = "neveropen" n = len(string) printChar(string, n)# This code is contributed by Ashutosh450 |
C#
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG {static int SIZE = 26;// Function to print the even frequency characters// in the order of their occurrencestatic 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 codepublic static void Main(String[] args){ String str = "neveropen"; int n = str.Length; printChar(str, n);}}// This code is contributed by Princi Singh |
Javascript
<script> // Javascript implementation of the approach let SIZE = 26; // Function to print the even 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 = new Array(SIZE); // Initialize all elements of freq[] to 0 for(let i = 0; i < freq.length; i++) { freq[i] = 0; } // Update the frequency of each character for (let i = 0; i < n; i++) { freq[str.charCodeAt(i) - 'a'.charCodeAt(0)]++; } // Traverse str character by character for (let i = 0; i < n; i++) { // If frequency of current character is even if (freq[str.charCodeAt(i) - 'a'.charCodeAt(0)] % 2 == 0) { document.write(str[i]); } } } let str = "neveropen"; let n = str.length; printChar(str, n); </script> |
Output:
neveropenneveropen
Time Complexity: O(n)
Auxiliary Space: O(1)
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