# Python3 Program to Count rotations divisible by 4

• Difficulty Level : Basic
• Last Updated : 09 Jun, 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.

## 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.`

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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