Data Modelling & AI Data Structure & Algorithm

Count even and odd digits in an Integer

31 July 2024

0

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) 
{ 
      // Initialize event_count and odd_count 
    int even_count = 0; 
    int odd_count = 0; 
        
    while (n > 0)  
    { 
        int rem = n % 10; 
          // if condition is true then increment even_count 
        if (rem % 2 == 0) { 
            even_count++; 
        } 
          // increment odd_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; 
      // Function call 
    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) 
{ 
      // Initialize event_count and odd_count 
    int even_count = 0; 
    int odd_count = 0; 
    while (n > 0)  
    { 
        int rem = n % 10; 
          // if condition is true then increment even_count 
        if (rem % 2 == 0){ 
            even_count++; 
        } 
        // increment odd_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; 
    // Function call 
    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): 
    # Initialize event_count and odd_count 
    even_count = 0
    odd_count = 0
    while (n > 0): 
        rem = n % 10
        # if condition is true then increment even_count 
        if (rem % 2 == 0): 
            even_count += 1
        # increment odd count 
        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; 
# Function call 
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) 
{ 
    // Initialize event_count and odd_count 
    int even_count = 0; 
    int odd_count = 0; 
    while (n > 0) { 
        int rem = n % 10; 
          // if condition is true then increment even_count 
        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; 
              // Function call 
            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) 
{ 
      // Initialize event_count and odd_count 
    $even_count = 0; 
    $odd_count = 0; 
    while ($n > 0)  
    { 
        $rem = $n % 10; 
          // if condition is true then increment even_count 
        if ($rem % 2 == 0) { 
            $even_count++; 
        } 
          // increment odd_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; 
// Function call 
$t = countEvenOdd($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 condition is true then increment even_count 
        if (rem % 2 == 0) { 
            even_count++;  
        } 
        // increment odd_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;  
    // Function call 
    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); 
      // Function cal 
    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; 
    // Function call 
    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)

This article is contributed by Dharmendra kumar. If you like neveropen and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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Count even and odd digits in an Integer

C++

Java

Python3

C#

PHP

Javascript

C++

Java

Python3

C#

PHP

Javascript

Method #3:Using typecasting(Simplified Approach):

C++

Java

Python3

C#

Javascript

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