# Java Program to Count rotations divisible by 4

• Last Updated : 06 Jan, 2022

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.

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

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