Count even and odd digits in an Integer

A certain number is given and the task is to count even digits and odd digits of the number and also of even digits are present even 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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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
  
# Drive 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. 
?> 

Output :

Even count : 2
Odd count : 5
YES

Another solution to solve this problem using 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. 
?> 

Output :

Even count : 1
Odd count : 2
NO

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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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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