Program to check if all characters have even frequency

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++

// 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

Java

// 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

Python3

# 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")

C#

// 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

Javascript

<script>

// JavaScript implementation of the above approach

function check(s)
{

    // creating a frequency array
    var freq = Array(26).fill(0);

    // Finding length of s
    var n = s.length;
    for (var 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 (var i = 0; i < 26; i++)
        if (freq[i] % 2 == 1)
        return false;
    return true;
}

// Driver Code
var s = "abaccaba";
check(s) ? document.write("Yes") :
           document.write("No");

</script>

Output

Yes

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(26) ⇒ O(1), no extra space is required, so it is a constant.

Method #2:Using built-in python functions.

Approach: We will scan the string and count the occurrence of all characters using the built-in Counter() function after that we traverse the counter list and check if the occurrences are even or not if there is any odd frequency character then we immediately print No. 

Note: This method is applicable for all types of characters

Python3

# Python implementation for
# the above approach

# importing Counter function
from collections import Counter

# Function to check if all
# elements occur even 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 odd count
        if (frequency[i] % 2 == 1):
            return False
    return True


# Driver code
s = "geeksforgeeksfor"
if(checkString(s)):
    print("Yes")
else:
    print("No")

# This code is contributed by vikkycirus

Output

Yes