# 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
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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

 `# Python 3 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

 ` `

Output:

```Rotations: 2
```

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

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Improved By : nitin mittal, jit_t

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