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# Java Program for Count rotations divisible by 8

Given a large positive number as string, count all rotations of the given number which are divisible by 8.

Examples:

Input: 8
Output: 1

Input: 40
Output: 1
Rotation: 40 is divisible by 8
04 is not divisible by 8

Input : 13502
Output : 0
No rotation is divisible by 8

Input : 43262488612
Output : 4

Approach: For large numbers it is difficult to rotate and divide each number by 8. Therefore, ‘divisibility by 8’ property is used which says that a number is divisible by 8 if the last 3 digits of the number is divisible by 8. Here we do not actually rotate the number and check last 8 digits for divisibility, instead we count consecutive sequence of 3 digits (in circular way) which are divisible by 8.

Illustration:

Consider a number 928160
Its rotations are 928160, 092816, 609281,
160928, 816092, 281609.
Now form consecutive sequence of 3-digits from
the original number 928160 as mentioned in the
approach.
3-digit: (9, 2, 8), (2, 8, 1), (8, 1, 6),
(1, 6, 0),(6, 0, 9), (0, 9, 2)
We can observe that the 3-digit number formed by
the these sets, i.e., 928, 281, 816, 160, 609, 092,
are present in the last 3 digits of some rotation.
Thus, checking divisibility of these 3-digit numbers
gives the required number of rotations.

## Java

 // Java program to count all// rotations divisible by 8import java.io.*; class GFG{    // function to count of all    // rotations divisible by 8    static int countRotationsDivBy8(String n)    {        int len = n.length();        int count = 0;             // For single digit number        if (len == 1) {            int oneDigit = n.charAt(0) - '0';            if (oneDigit % 8 == 0)                return 1;            return 0;        }             // For two-digit numbers        // (considering all pairs)        if (len == 2) {                 // first pair            int first = (n.charAt(0) - '0') *                        10 + (n.charAt(1) - '0');                 // second pair            int second = (n.charAt(1) - '0') *                         10 + (n.charAt(0) - '0');                 if (first % 8 == 0)                count++;            if (second % 8 == 0)                count++;            return count;        }             // considering all three-digit sequences        int threeDigit;        for (int i = 0; i < (len - 2); i++)        {            threeDigit = (n.charAt(i) - '0') * 100 +                        (n.charAt(i + 1) - '0') * 10 +                        (n.charAt(i + 2) - '0');            if (threeDigit % 8 == 0)                count++;        }             // Considering the number formed by the        // last digit and the first two digits        threeDigit = (n.charAt(len - 1) - '0') * 100 +                    (n.charAt(0) - '0') * 10 +                    (n.charAt(1) - '0');             if (threeDigit % 8 == 0)            count++;             // Considering the number formed by the last        // two digits and the first digit        threeDigit = (n.charAt(len - 2) - '0') * 100 +                    (n.charAt(len - 1) - '0') * 10 +                    (n.charAt(0) - '0');        if (threeDigit % 8 == 0)            count++;             // required count of rotations        return count;    }         // Driver program    public static void main (String[] args)    {        String n = "43262488612";        System.out.println( "Rotations: "                       +countRotationsDivBy8(n));             }} // This code is contributed by vt_m.

Output:

Rotations: 4

Time Complexity : O(n), where n is the number of digits in input number.
Auxiliary Space: O(1)

Please refer complete article on Count rotations divisible by 8 for more details!

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