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 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

Time Complexity: O(n)

Auxiliary Space: O(1)

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

Time Complexity: O(n)

Auxiliary Space: O(1)

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:

C++

// C++ implementation of above approach
#include<bits/stdc++.h>
using namespace std;

string getResult(int n)
{

  // Converting integer to String
  string st = to_string(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
int main(){

  int n = 77788;

  // Passing this number to get result function
  cout<<getResult(n)<<endl;

}

// This code is contributed by shinjanpatra.

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

Time Complexity: O(n)

Auxiliary Space: O(1)

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