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# Count rotations which are divisible by 10

• Difficulty Level : Easy
• Last Updated : 18 Nov, 2021

Given a number N, the task is to count all the rotations of the given number which are divisible by 10.
Examples:

Input: N = 10203
Output:
Explanation:
There are 5 rotations possible for the given number. They are: 02031, 20310, 03102, 31020, 10203
Out of these rotations, only 20310 and 31020 are divisible by 10. So 2 is the output.
Input: N = 135
Output:

Naive Approach: The naive approach for this problem is to form all the possible rotations. It is known that for a number of size K, the number of possible rotations for this number N is K. Therefore, find all the rotations and for every rotation, check if the number is divisible by 10 or not. The time complexity for this approach is quadratic.
Efficient Approach: The efficient approach lies behind the concept that in order to check whether a number is divisible by 10 or not, we simply check if the last digit is 0. So, the idea is to simply iterate over the given number and find the count of 0’s. If the count of 0’s is F, then clearly, F out of K rotations will have 0 at the end of the given number N
Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the``// count of rotations which are``// divisible by 10` `#include ``using` `namespace` `std;` `// Function to return the count of``// all the rotations which are``// divisible by 10.``int` `countRotation(``int` `n)``{``    ``int` `count = 0;` `    ``// Loop to iterate through the``    ``// number``    ``do` `{``        ``int` `digit = n % 10;` `        ``// If the last digit is 0,``        ``// then increment the count``        ``if` `(digit == 0)``            ``count++;``        ``n = n / 10;``    ``} ``while` `(n != 0);` `    ``return` `count;``}` `// Driver code``int` `main()``{``    ``int` `n = 10203;``    ``cout << countRotation(n);``}`

## C#

 `// CSharp implementation to find the``// count of rotations which are``// divisible by 10` `using` `System;``class` `Solution {` `    ``// Function to return the count``    ``// of all rotations which are``    ``// divisible by 10.``    ``static` `int` `countRotation(``int` `n)``    ``{``        ``int` `count = 0;` `        ``// Loop to iterate through the``        ``// number``        ``do` `{``            ``int` `digit = n % 10;` `            ``// If the last digit is 0,``            ``// then increment the count``            ``if` `(digit % 2 == 0)``                ``count++;``            ``n = n / 10;``        ``} ``while` `(n != 0);` `        ``return` `count;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 10203;``        ``Console.Write(countRotation(n));``    ``}``}`

## Java

 `// Java implementation to find the``// count of rotations which are``// divisible by 10` `class` `GFG {` `    ``// Function to return the count``    ``// of all rotations which are``    ``// divisible by 10.``    ``static` `int` `countRotation(``int` `n)``    ``{``        ``int` `count = ``0``;` `        ``// Loop to iterate through the``        ``// number``        ``do` `{``            ``int` `digit = n % ``10``;` `            ``// If the last digit is 0,``            ``// then increment the count``            ``if` `(digit == ``0``)``                ``count++;``            ``n = n / ``10``;``        ``} ``while` `(n != ``0``);` `        ``return` `count;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``10203``;` `        ``System.out.println(countRotation(n));``    ``}``}`

## Python

 `# Python3 implementation to find the``# count of rotations which are``# divisible by 10` `# Function to return the count of``# all rotations which are divisible``# by 10.``def` `countRotation(n):``    ``count ``=` `0``;` `    ``# Loop to iterate through the``    ``# number``    ``while` `n > ``0``:``        ``digit ``=` `n ``%` `10` `        ``# If the last digit is 0,``        ``# then increment the count``        ``if``(digit ``%` `2` `=``=` `0``):``            ``count ``=` `count ``+` `1``        ``n ``=` `int``(n ``/` `10``)``    ` `    ``return` `count;   ``  ` `# Driver code ``if` `__name__ ``=``=` `"__main__"` `:``  ` `    ``n ``=` `10203``; ``    ``print``(countRotation(n)); `

## Javascript

 ``

Output:

`2`

Time Complexity: O(log10N), where N is the length of the number.

Auxiliary Space: O(1)

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