Count rotations of N which are Odd and Even

Difficulty Level : Basic
Last Updated : 26 Mar, 2021

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++

// 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);
}

Java

// 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);
    }
}

Python3

# 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

C#

// 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);
    }
}

PHP

<?php
// PHP implementation of the above approach
 
// Function to count of all rotations
// which are odd and even
function countOddRotations($n)
{
    $odd_count = 0;
    $even_count = 0;
    do {
        $digit = $n % 10;
        if ($digit % 2 == 1)
            $odd_count++;
        else
            $even_count++;
        $n = (int)($n / 10);
    } while ($n != 0);
 
    echo "Odd = ", $odd_count, "\n";
    echo "Even = ", $even_count, "\n";
}
 
// Driver Code
$n = 1234;
countOddRotations($n);
 
// This code is contributed by ajit..
?>

Javascript

<script>
 
// Javascript implementation of the above approach
 
// Function to count of all rotations
// which are odd and even
function countOddRotations(n)
{
    var odd_count = 0, even_count = 0;
    do {
        var digit = n % 10;
        if (digit % 2 == 1)
            odd_count++;
        else
            even_count++;
        n = parseInt(n / 10);
    } while (n != 0);
 
    document.write("Odd = " + odd_count + "<br>");
    document.write("Even = " + even_count + "<br>");
}
 
// Driver Code
var n = 1234;
countOddRotations(n);
 
// This code is contributed by rutvik_56.
 
</script>

Output:

Odd = 2
Even = 2

Time Complexity: O(n)

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