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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 ``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

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## Javascript

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Output:
```Odd = 2
Even = 2```

Time Complexity: O(n)

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