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)
{
      // 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;
}

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.

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:

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)