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# Count rotations divisible by 4

• Difficulty Level : Easy
• Last Updated : 08 Jul, 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 ```
Recommended Practice

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.

## C++

 `// C++ program to count all rotation divisible` `// by 4.` `#include ` `using` `namespace` `std;`   `// Returns count of all rotations divisible` `// by 4` `int` `countRotations(string n)` `{` `    ``int` `len = n.length();`   `    ``// For single digit number` `    ``if` `(len == 1)` `    ``{` `        ``int` `oneDigit = n.at(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.at(i)-``'0'``)*10 + (n.at(i+1)-``'0'``);` `        ``if` `(twoDigit%4 == 0)` `            ``count++;` `    ``}`   `    ``// Considering the number formed by the pair of` `    ``// last digit and 1st digit` `    ``twoDigit = (n.at(len-1)-``'0'``)*10 + (n.at(0)-``'0'``);` `    ``if` `(twoDigit%4 == 0)` `        ``count++;`   `    ``return` `count;` `}`   `//Driver program` `int` `main()` `{` `    ``string n = ``"4834"``;` `    ``cout << ``"Rotations: "` `<< countRotations(n) << endl;` `    ``return` `0;` `}`

## 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.`

## Python3

 `# Python3 program to count` `# all rotation divisible` `# by 4.`   `# Returns count of all` `# rotations divisible` `# by 4` `def` `countRotations(n) :`   `    ``l ``=` `len``(n)`   `    ``# For single digit number` `    ``if` `(l ``=``=` `1``) :` `        ``oneDigit ``=` `(``int``)(n[``0``])` `        `  `        ``if` `(oneDigit ``%` `4` `=``=` `0``) :` `            ``return` `1` `        ``return` `0` `    `  `    `  `    ``# At-least 2 digit number` `    ``# (considering all pairs)` `    ``count ``=` `0` `    ``for` `i ``in` `range``(``0``, l ``-` `1``) :` `        ``twoDigit ``=` `(``int``)(n[i]) ``*` `10` `+` `(``int``)(n[i ``+` `1``])` `        `  `        ``if` `(twoDigit ``%` `4` `=``=` `0``) :` `            ``count ``=` `count ``+` `1` `            `  `    ``# Considering the number` `    ``# formed by the pair of` `    ``# last digit and 1st digit` `    ``twoDigit ``=` `(``int``)(n[l ``-` `1``]) ``*` `10` `+` `(``int``)(n[``0``])` `    ``if` `(twoDigit ``%` `4` `=``=` `0``) :` `        ``count ``=` `count ``+` `1`   `    ``return` `count`   `# Driver program` `n ``=` `"4834"` `print``(``"Rotations: "` `,` `    ``countRotations(n))`   `# This code is contributed by Nikita tiwari.`

## C#

 `// C# program to count all rotation` `// divisible by 4.` `using` `System;`   `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 - ``'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[i] - ``'0'``) * 10 +` `                          ``(n[i + 1] - ``'0'``);` `    `  `            ``if` `(twoDigit % 4 == 0)` `                ``count++;` `        ``}` `    `  `        ``// Considering the number` `        ``// formed by the pair of` `        ``// last digit and 1st digit` `        ``twoDigit = (n[len - 1] - ``'0'``) * 10 +` `                               ``(n - ``'0'``);`   `        ``if` `(twoDigit % 4 == 0)` `            ``count++;` `    `  `        ``return` `count;` `    ``}` `    `  `    ``//Driver program` `    ``public` `static` `void` `Main()` `    ``{` `        ``String n = ``"4834"``;` `        ``Console.Write(``"Rotations: "` `+` `                    ``countRotations(n));` `    ``}` `}`   `// This code is contributed by nitin mittal.`

## PHP

 ``

## Javascript

 ``

Output

`Rotations: 2`

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

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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