Count rotations of N which are Odd and Even

Given a number n, the task is to count all rotations of the given number which are odd and even.

Examples:

Input: n = 1234
Output: Odd = 2, Even = 2
Total rotations: 1234, 2341, 3412, 4123
Odd rotations: 2341 and 4123
Even rotations: 1234 and 3412

Input: n = 246
Output: Odd = 0, Even = 3

Efficient Approach: For large numbers, it is difficult to rotate and check whether it is odd or not for every rotation. Hence, in this approach, check the count of odd digits and even digits present in the number. These will be the answer to this problem.

Below is the implementation of the above approach:

Implementation:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to count of all rotations
// which are odd and even
void countOddRotations(int n)
{
    int odd_count = 0, even_count = 0;
    do {
        int digit = n % 10;
        if (digit % 2 == 1)
            odd_count++;
        else
            even_count++;
        n = n / 10;
    } while (n != 0);
  
    cout << "Odd = " << odd_count << endl;
    cout << "Even = " << even_count << endl;
}
  
// Driver Code
int main()
{
    int n = 1234;
    countOddRotations(n);
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the above approach
  
class Solution {
  
    // Function to count of all rotations
    // which are odd and even
    static void countOddRotations(int n)
    {
        int odd_count = 0, even_count = 0;
        do {
            int digit = n % 10;
            if (digit % 2 == 1)
                odd_count++;
            else
                even_count++;
            n = n / 10;
        } while (n != 0);
  
        System.out.println("Odd = " + odd_count);
        System.out.println("Even = " + even_count);
    }
  
    public static void main(String[] args)
    {
        int n = 1234;
        countOddRotations(n);
    }
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python implementation of the above approach
  
# Function to count of all rotations
# which are odd and even
def countOddRotations(n):
    odd_count = 0; even_count = 0
    while n != 0:
        digit = n % 10
        if digit % 2 == 0:
            odd_count += 1
        else:
            even_count += 1
        n = n//10
    print("Odd =", odd_count)
    print("Even =", even_count)
  
# Driver code
n = 1234
countOddRotations(n)
  
# This code is contributed by Shrikant13

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// CSharp implementation of the above approach
  
using System;
class Solution {
  
    // Function to count of all rotations
    // which are odd and even
    static void countOddRotations(int n)
    {
        int odd_count = 0, even_count = 0;
        do {
            int digit = n % 10;
            if (digit % 2 == 1)
                odd_count++;
            else
                even_count++;
            n = n / 10;
        } while (n != 0);
  
        Console.WriteLine("Odd = " + odd_count);
        Console.WriteLine("Even = " + even_count);
    }
  
    public static void Main()
    {
        int n = 1234;
        countOddRotations(n);
    }
}

chevron_right


PHP

Output:

Odd = 2
Even = 2

Time Complexity: O(n)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : shrikanth13, jit_t