Related Articles

Related Articles

Program to check if all characters have even frequency
  • Last Updated : 12 Aug, 2019

Given a string S consisting only of lowercase letters check if the string has all characters appearing even times.

Examples:

Input : abaccaba
Output : Yes
Explanation: ‘a’ occurs four times, ‘b’ occurs twice, ‘c’ occurs twice and the other letters occur zero times.

Input : hthth
Output : No

Approach:
We will go through the string and count the occurrence of all characters after that we will check if the occurrences are even or not if there is any odd frequency character then we immediately print No.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the above approach
#include <iostream>
using namespace std;
  
bool check(string s)
{
      
    // creating a frequency array
    int freq[26] = {0};
      
    // Finding length of s
    int n = s.length();
    for (int i = 0; i < s.length(); i++)
      
    // counting frequency of all characters
        freq[s[i] - 97]++;
      
    // checking if any odd frequency 
    // is there or not
    for (int i = 0; i < 26; i++)
        if (freq[i] % 2 == 1)
        return false;
    return true;
}
  
// Driver Code
int main() 
{
    string s = "abaccaba";
    check(s) ? cout << "Yes" << endl : 
               cout << "No" << endl;
    return 0;
}
  
// This code is contributed by
// sanjeev2552

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the above approach
class GFG 
{
    static boolean check(String s) 
    {
  
        // creating a frequency array
        int[] freq = new int[26];
  
        // Finding length of s
        int n = s.length();
  
        // counting frequency of all characters
        for (int i = 0; i < s.length(); i++)
        {
            freq[(s.charAt(i)) - 97] += 1;
        }
  
        // checking if any odd frequency
        // is there or not
        for (int i = 0; i < freq.length; i++) 
        {
            if (freq[i] % 2 == 1
            {
                return false;
            }
        }
        return true;
    }
  
    // Driver Code
    public static void main(String[] args) 
    {
        String s = "abaccaba";
        if (check(s)) 
        {
            System.out.println("Yes");
        
        else 
        {
            System.out.println("No");
        }
    }
}
  
// This code is contributed by Rajput-Ji

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python implementation of the above approach
def check(s):
  
    # creating a frequency array
    freq =[0]*26
  
    # Finding length of s
    n = len(s)
  
    for i in range(n):
  
        # counting frequency of all characters
        freq[ord(s[i])-97]+= 1
  
    for i in range(26):
  
        # checking if any odd frequency
        # is there or not
        if (freq[i]% 2 == 1):
            return False
    return True
  
# Driver code
s ="abaccaba"
if(check(s)):
    print("Yes")
else:
    print("No")

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
      
class GFG 
{
    static Boolean check(String s) 
    {
  
        // creating a frequency array
        int[] freq = new int[26];
  
        // Finding length of s
        int n = s.Length;
  
        // counting frequency of all characters
        for (int i = 0; i < s.Length; i++)
        {
            freq[(s[i]) - 97] += 1;
        }
  
        // checking if any odd frequency
        // is there or not
        for (int i = 0; i < freq.Length; i++) 
        {
            if (freq[i] % 2 == 1) 
            {
                return false;
            }
        }
        return true;
    }
  
    // Driver Code
    public static void Main(String[] args) 
    {
        String s = "abaccaba";
        if (check(s)) 
        {
            Console.WriteLine("Yes");
        
        else
        {
            Console.WriteLine("No");
        }
    }
}
  
// This code is contributed by PrinciRaj1992 

chevron_right


Output:

Yes

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up
Recommended Articles
Page :