# 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!

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