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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 8``import` `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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