Java Program to Count rotations divisible by 4

Given a large positive number as string, count all rotations of the given number which are divisible by 4. 

Examples:

Input: 8
Output: 1

Input: 20
Output: 1
Rotation: 20 is divisible by 4
          02 is not divisible by 4

Input : 13502
Output : 0
No rotation is divisible by 4

Input : 43292816
Output : 5
5 rotations are : 43292816, 16432928, 81643292
                  92816432, 32928164 

For large numbers it is difficult to rotate and divide each number by 4. Therefore, ‘divisibility by 4’ property is used which says that a number is divisible by 4 if the last 2 digits of the number is divisible by 4. Here we do not actually rotate the number and check last 2 digits for divisibility, instead we count consecutive pairs (in circular way) which are divisible by 4. 

Illustration:

Consider a number 928160
Its rotations are 928160, 092816, 609281, 160928, 
    816092, 281609.
Now form pairs from the original number 928160
as mentioned in the approach.
Pairs: (9,2), (2,8), (8,1), (1,6), 
         (6,0), (0,9)
We can observe that the 2-digit number formed by the these 
pairs, i.e., 92, 28, 81, 16, 60, 09, are present in the last
2 digits of some rotation.
Thus, checking divisibility of these pairs gives the required
number of rotations. 

Note: A single digit number can directly
be checked for divisibility.

Below is the implementation of the approach. 

Java

// Java program to count
// all rotation divisible
// by 4.
import java.io.*;

class GFG {

    // Returns count of all
    // rotations divisible
    // by 4
    static int countRotations(String n)
    {
        int len = n.length();

        // For single digit number
        if (len == 1)
        {
          int oneDigit = n.charAt(0)-'0';

          if (oneDigit % 4 == 0)
              return 1;

          return 0;
        }

        // At-least 2 digit
        // number (considering all
        // pairs)
        int twoDigit, count = 0;
        for (int i = 0; i < (len-1); i++)
        {
          twoDigit = (n.charAt(i)-'0') * 10 +
                     (n.charAt(i+1)-'0');

          if (twoDigit%4 == 0)
              count++;
        }

        // Considering the number
        // formed by the pair of
        // last digit and 1st digit
        twoDigit = (n.charAt(len-1)-'0') * 10 +
                   (n.charAt(0)-'0');

        if (twoDigit%4 == 0)
            count++;

        return count;
    }

    //Driver program
    public static void main(String args[])
    {
        String n = "4834";
        System.out.println("Rotations: " +
                          countRotations(n));
    }
}

// This code is contributed by Nikita tiwari.

Output:

Rotations: 2

Time Complexity : O(n) where n is number of digits in input number.

Auxiliary Space: O(1)

Please refer complete article on Count rotations divisible by 4 for more details!