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# Count rotations divisible by 4

• Difficulty Level : Easy
• Last Updated : 09 Apr, 2021

Given a large positive number as string, count all rotations of the given number which are divisible by 4.

Examples:

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```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.

## 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.

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