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Count even and odd digits in an Integer

  • Difficulty Level : Easy
  • Last Updated : 29 May, 2021

A certain number is given and the task is to count even digits and odd digits of the number and also even digits are present even a number of times and, similarly, for odd numbers. 

Print Yes If:
   If number contains even digits even number of time
   Odd digits odd number of times
Else 
   Print No

Examples :

Input : 22233
Output : NO
         count_even_digits = 3
         count_odd_digits = 2
         In this number even digits occur odd number of times and odd 
         digits occur even number of times so its print NO.

Input : 44555
Output : YES
        count_even_digits = 2
        count_odd_digits = 3
        In this number even digits occur even number of times and odd 
        digits occur odd number of times so its print YES.

Efficient solution for calculating even and odd digits in a number.  

C++




// C++ program to count
// even and odd digits
// in a given number
#include <iostream>
using namespace std;
 
// Function to count digits
int countEvenOdd(int n)
{
    int even_count = 0;
    int odd_count = 0;
    while (n > 0)
    {
        int rem = n % 10;
        if (rem % 2 == 0)
            even_count++;
        else
            odd_count++;
        n = n / 10;
    }
    cout << "Even count : "
         << even_count;
    cout << "\nOdd count : "
         << odd_count;
    if (even_count % 2 == 0 &&
        odd_count % 2 != 0)
        return 1;
    else
        return 0;
}
 
// Driver Code
int main()
{
    int n;
    n = 2335453;
    int t = countEvenOdd(n);
    if (t == 1)
        cout << "\nYES" << endl;
    else
        cout << "\nNO" << endl;
    return 0;
}

Java




// Java program to count
// even and odd digits
// in a given number
 
import java.io.*;
 
class GFG
{
     
// Function to count digits
static int countEvenOdd(int n)
{
    int even_count = 0;
    int odd_count = 0;
    while (n > 0)
    {
        int rem = n % 10;
        if (rem % 2 == 0)
            even_count++;
        else
            odd_count++;
        n = n / 10;
    }
    System.out.println ( "Even count : " +
                              even_count);
    System.out.println ( "Odd count : " +
                              odd_count);
    if (even_count % 2 == 0 &&
         odd_count % 2 != 0)
        return 1;
    else
        return 0;
}
 
    // Driver Code
    public static void main (String[] args)
    {
    int n;
    n = 2335453;
    int t = countEvenOdd(n);
     
    if (t == 1)
        System.out.println ( "YES" );
    else
        System.out.println( "NO") ;
         
    }
}

Python3




# python program to count even and
# odd digits in a given number
 
# Function to count digits
def countEvenOdd(n):
     
    even_count = 0
    odd_count = 0
    while (n > 0):
        rem = n % 10
        if (rem % 2 == 0):
            even_count += 1
        else:
            odd_count += 1
             
        n = int(n / 10)
     
    print( "Even count : " , even_count)
    print("\nOdd count : " , odd_count)
    if (even_count % 2 == 0 and
                    odd_count % 2 != 0):
        return 1
    else:
        return 0
 
# Driver code
n = 2335453;
t = countEvenOdd(n);
 
if (t == 1):
    print("YES")
else:
    print("NO")
 
# This code is contributed by Sam007.

C#




// C# program to count even and
// odd digits in a given number
using System;
 
class GFG {
     
// Function to count digits
static int countEvenOdd(int n)
{
    int even_count = 0;
    int odd_count = 0;
    while (n > 0) {
        int rem = n % 10;
        if (rem % 2 == 0)
            even_count++;
        else
            odd_count++;
        n = n / 10;
    }
     
    Console.WriteLine("Even count : " +
                       even_count);
    Console.WriteLine("Odd count : " +
                       odd_count);
    if (even_count % 2 == 0 &&
        odd_count % 2 != 0)
        return 1;
    else
        return 0;
}
 
    // Driver Code
    public static void Main ()
    {
            int n;
            n = 2335453;
            int t = countEvenOdd(n);
            if (t == 1)
                Console.WriteLine ("YES");
            else
                Console.WriteLine("NO") ;
             
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to count
// even and odd digits
// in a given number
 
// Function to count digits
function countEvenOdd($n)
{
    $even_count = 0;
    $odd_count = 0;
    while ($n > 0)
    {
        $rem = $n % 10;
        if ($rem % 2 == 0)
            $even_count++;
        else
            $odd_count++;
        $n = (int)($n / 10);
    }
    echo("Even count : " .
             $even_count);
    echo("\nOdd count : " .
               $odd_count);
    if ($even_count % 2 == 0 &&
        $odd_count % 2 != 0)
        return 1;
    else
        return 0;
}
 
// Driver code
$n = 2335453;
$t = countEvenOdd($n);
if ($t == 1)
    echo("\nYES");
else
    echo("\nNO");
 
// This code is contributed by Ajit.
?>

Javascript




<script>
 
/// Javascript program to count
// even and odd digits
// in a given number
 
// Function to count digits
function countEvenOdd(n)
{
    let even_count = 0;
    let odd_count = 0;
    while (n > 0)
    {
        let rem = n % 10;
        if (rem % 2 == 0)
            even_count++;
        else
            odd_count++;
        n = Math.floor(n / 10);
    }
    document.write("Even count : "
        + even_count);
    document.write("<br>" + "Odd count : "
        + odd_count);
    if (even_count % 2 == 0 &&
        odd_count % 2 != 0)
        return 1;
    else
        return 0;
}
 
// Driver Code
 
    let n;
    n = 2335453;
    let t = countEvenOdd(n);
    if (t == 1)
        document.write("<br>" + "YES" + "<br>");
    else
        document.write("<br>"+"NO" + "<br>");
 
// This code is contributed by Mayank Tyagi
 
</script>
Output
Even count : 2
Odd count : 5
YES

Another solution to solve this problem is a character array or string.



C++




// C++ program to count
// even and odd digits
// in a given number
// using char array
#include <bits/stdc++.h>
using namespace std;
 
// Function to count digits
int countEvenOdd(char num[],
                 int n)
{
    int even_count = 0;
    int odd_count = 0;
    for (int i = 0; i < n; i++)
    {
        int x = num[i] - 48;
        if (x % 2 == 0)
            even_count++;
        else
            odd_count++;
    }
    cout << "Even count : "
         << even_count;
    cout << "\nOdd count : "
         << odd_count;
 
    if (even_count % 2 == 0 &&
        odd_count % 2 != 0)
        return 1;
    else
        return 0;
}
 
// Driver Code
int main()
{
    char num[18] = { 1, 2, 3 };
 
    int n = strlen(num);
    int t = countEvenOdd(num, n);
    if (t == 1)
        cout << "\nYES" << endl;
    else
        cout << "\nNO" << endl;
    return 0;
}

Java




// Java program to count
// even and odd digits
// in a given number
// using char array
 
import java.io.*;
 
 
class GFG
{
     
// Function to count digits
static int countEvenOdd(char num[],
                        int n)
{
    int even_count = 0;
    int odd_count = 0;
    for (int i = 0; i < n; i++)
    {
        int x = num[i] - 48;
        if (x % 2 == 0)
            even_count++;
        else
            odd_count++;
    }
 
    System.out.println ("Even count : " +
                         even_count);
    System.out.println( "Odd count : " +
                         odd_count);
 
    if (even_count % 2 == 0 &&
        odd_count % 2 != 0)
        return 1;
    else
        return 0;
}
 
    // Driver Code
    public static void main (String[] args)
    {
        char num[] = { 1, 2, 3 };
 
    int n = num.length;
    int t = countEvenOdd(num, n);
    if (t == 1)
        System.out.println("YES") ;
    else
        System.out.println("NO") ;
    }
}
 
// This code is contributed by vt_m

Python3




# Python3 program to count
# even and odd digits
# in a given number
# using char array
 
# Function to count digits
def countEvenOdd(num, n):
    even_count = 0;
    odd_count = 0;
    num=list(str(num))
    for i in num:
        if i in ('0','2','4','6','8'):
            even_count+=1
        else:
            odd_count+=1
    print("Even count : ",
              even_count);
    print("Odd count : ",
              odd_count);
    if (even_count % 2 == 0 and
        odd_count % 2 != 0):
        return 1;
    else:
        return 0;
 
# Driver Code
num = (1, 2, 3);
n = len(num);
t = countEvenOdd(num, n);
 
if t == 1:
    print("YES");
else:
    print("NO");
     
# This code is contributed by mits.

C#




// C# program to count
// even and odd digits
// in a given number
// using char array
using System;
 
class GFG
{
     
    // Function to count digits
    static int countEvenOdd(char []num,
                            int n)
    {
        int even_count = 0;
        int odd_count = 0;
        for (int i = 0; i < n; i++)
        {
            int x = num[i] - 48;
            if (x % 2 == 0)
                even_count++;
            else
                odd_count++;
        }
     
        Console.WriteLine("Even count : " +
                               even_count);
                             
        Console.WriteLine( "Odd count : " +
                                odd_count);
     
        if (even_count % 2 == 0 &&
            odd_count % 2 != 0)
            return 1;
        else
            return 0;
    }
 
    // Driver code
    public static void Main ()
    {
        char [] num = { '1', '2', '3' };
     
        int n = num.Length;
        int t = countEvenOdd(num, n);
         
        if (t == 1)
            Console.WriteLine("YES") ;
        else
            Console.WriteLine("NO") ;
    }
}
 
// This code is contributed by Sam007.

PHP




<?php
// PHP program to count
// even and odd digits
// in a given number
// using char array
 
 
// Function to count digits
function countEvenOdd($num, $n)
{
     
    $even_count = 0;
    $odd_count = 0;
    for ($i = 0; $i < $n; $i++)
    {
        $x = $num[$i] - 48;
        if ($x % 2 == 0)
            $even_count++;
        else
            $odd_count++;
    }
    echo "Even count : " .
          $even_count;
    echo "\nOdd count : " .
            $odd_count;
 
    if ($even_count % 2 == 0 &&
        $odd_count % 2 != 0)
        return 1;
    else
        return 0;
}
 
    // Driver Code
    $num = array( 1, 2, 3 );
 
    $n = strlen(num);
    $t = countEvenOdd($num, $n);
    if ($t == 1)
        echo "\nYES" ,"\n";
    else
        echo "\nNO" ,"\n";
 
// This code is contributed by ajit.
?>

Javascript




<script>
 
// Javascript program to count
// even and odd digits
// in a given number
// using char array
 
// Function to count digits
function countEvenOdd(num, n)
{
    even_count = 0;
    odd_count = 0;
     
    for(var i = 0; i < n; i++)
    {
        x = num[i] - 48;
        if (x % 2 == 0)
            even_count++;
        else
            odd_count++;
    }
     
    document.write("Even count : " +
                   even_count);
    document.write("<br>Odd count : " +
                   odd_count);
  
    if (even_count % 2 == 0 &&
        odd_count % 2 != 0)
        return 1;
    else
        return 0;
}
 
// Driver code
var num = [ 1, 2, 3 ];
n = num.length;
t = countEvenOdd(num, n);
 
if (t == 1)
    document.write("<br>YES <br>");
else
    document.write("<br>NO <br>");
     
// This code is contributed by akshitsaxenaa09
 
</script>
Output
Even count : 1
Odd count : 2
NO

Method #3:Using typecasting(Simplified Approach):

  • We have to convert the given number to a string by taking a new variable.
  • Traverse the string, convert each element to an integer.
  • If the character(digit) is even, then the increased count
  • Else increment the odd count.
  • If the even count is even and the odd count is odd, then print Yes.
  • Else print no.

Below is the implementation of the above approach:

Java




// Java implementation of above approach
class GFG{
     
static String getResult(int n)
{
     
    // Converting integer to String
    String st = String.valueOf(n);
    int even_count = 0;
    int odd_count = 0;
     
    // Looping  till length of String
    for(int i = 0; i < st.length(); i++)
    {
        if ((st.charAt(i) % 2) == 0)
         
            // Digit is even so increment even count
            even_count += 1;
        else
            odd_count += 1;
    }
     
    // Checking even count is even and
    // odd count is odd
    if (even_count % 2 == 0 &&
         odd_count % 2 != 0)
        return "Yes";
    else
        return "no";
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 77788;
     
    // Passing this number to get result function
    System.out.println(getResult(n));
 
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python implementation of above approach
def getResult(n):
   
    # Converting integer to string
    st = str(n)
    even_count = 0
    odd_count = 0
     
    # Looping  till length of string
    for i in range(len(st)):
       
        if((int(st[i]) % 2) == 0):
             
            # digit is even so increment even count
            even_count += 1
        else:
            odd_count += 1
 
    # Checking even count is even and odd count is odd
    if(even_count % 2 == 0 and odd_count % 2 != 0):
        return 'Yes'
    else:
        return 'no'
 
 
# Driver Code
n = 77788
 
# passing this number to get result function
print(getResult(n))
 
# this code is contributed by vikkycirus

C#




// C# implementation of above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
static String getResult(int n)
{
     
    // Converting integer to String
    String st = String.Join("",n);
    int even_count = 0;
    int odd_count = 0;
     
    // Looping  till length of String
    for(int i = 0; i < st.Length; i++)
    {
        if ((st[i] % 2) == 0)
         
            // Digit is even so increment even count
            even_count += 1;
        else
            odd_count += 1;
    }
     
    // Checking even count is even and
    // odd count is odd
    if (even_count % 2 == 0 &&
         odd_count % 2 != 0)
        return "Yes";
    else
        return "no";
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 77788;
     
    // Passing this number to get result function
    Console.WriteLine(getResult(n));
 
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
// Javascript implementation of above approach
 
 
function getResult(n)
{
     
    // Converting integer to String
    var st = n.toString();
    var even_count = 0;
    var odd_count = 0;
     
    // Looping  till length of String
    for(var i = 0; i < st.length; i++)
    {
        if ((st.charAt(i) % 2) == 0)
         
            // Digit is even so increment even count
            even_count += 1;
        else
            odd_count += 1;
    }
     
    // Checking even count is even and
    // odd count is odd
    if (even_count % 2 == 0 &&
         odd_count % 2 != 0)
        return "Yes";
    else
        return "no";
}
 
// Driver Code
 
    var n = 77788;
     
    // Passing this number to get result function
    document.write(getResult(n));
     
</script>
Output
Yes

This article is contributed by Dharmendra kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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