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# Python3 Program for Count rotations divisible by 8

• Last Updated : 09 Jun, 2022

Given a large positive number as string, count all rotations of the given number which are divisible by 8.

Examples:

```Input: 8
Output: 1

Input: 40
Output: 1
Rotation: 40 is divisible by 8
04 is not divisible by 8

Input : 13502
Output : 0
No rotation is divisible by 8

Input : 43262488612
Output : 4```

Approach: For large numbers it is difficult to rotate and divide each number by 8. Therefore, ‘divisibility by 8’ property is used which says that a number is divisible by 8 if the last 3 digits of the number is divisible by 8. Here we do not actually rotate the number and check last 8 digits for divisibility, instead we count consecutive sequence of 3 digits (in circular way) which are divisible by 8.

Illustration:

```Consider a number 928160
Its rotations are 928160, 092816, 609281,
160928, 816092, 281609.
Now form consecutive sequence of 3-digits from
the original number 928160 as mentioned in the
approach.
3-digit: (9, 2, 8), (2, 8, 1), (8, 1, 6),
(1, 6, 0),(6, 0, 9), (0, 9, 2)
We can observe that the 3-digit number formed by
the these sets, i.e., 928, 281, 816, 160, 609, 092,
are present in the last 3 digits of some rotation.
Thus, checking divisibility of these 3-digit numbers
gives the required number of rotations. ```

## Python3

 `# Python3 program to count all``# rotations divisible by 8` `# function to count of all``# rotations divisible by 8``def` `countRotationsDivBy8(n):``    ``l ``=` `len``(n)``    ``count ``=` `0` `    ``# For single digit number``    ``if` `(l ``=``=` `1``):``        ``oneDigit ``=` `int``(n[``0``])``        ``if` `(oneDigit ``%` `8` `=``=` `0``):``            ``return` `1``        ``return` `0` `    ``# For two-digit numbers``    ``# (considering all pairs)``    ``if` `(l ``=``=` `2``):` `        ``# first pair``        ``first ``=` `int``(n[``0``]) ``*` `10` `+` `int``(n[``1``])` `        ``# second pair``        ``second ``=` `int``(n[``1``]) ``*` `10` `+` `int``(n[``0``])` `        ``if` `(first ``%` `8` `=``=` `0``):``            ``count``+``=``1``        ``if` `(second ``%` `8` `=``=` `0``):``            ``count``+``=``1``        ``return` `count` `    ``# considering all``    ``# three-digit sequences``    ``threeDigit``=``0``    ``for` `i ``in` `range``(``0``,(l ``-` `2``)):``        ``threeDigit ``=` `(``int``(n[i]) ``*` `100` `+``                     ``int``(n[i ``+` `1``]) ``*` `10` `+``                     ``int``(n[i ``+` `2``]))``        ``if` `(threeDigit ``%` `8` `=``=` `0``):``            ``count``+``=``1` `    ``# Considering the number``    ``# formed by the last digit``    ``# and the first two digits``    ``threeDigit ``=` `(``int``(n[l ``-` `1``]) ``*` `100` `+``                 ``int``(n[``0``]) ``*` `10` `+``                 ``int``(n[``1``]))` `    ``if` `(threeDigit ``%` `8` `=``=` `0``):``        ``count``+``=``1` `    ``# Considering the number``    ``# formed by the last two``    ``# digits and the first digit``    ``threeDigit ``=` `(``int``(n[l ``-` `2``]) ``*` `100` `+``                 ``int``(n[l ``-` `1``]) ``*` `10` `+``                 ``int``(n[``0``]))``    ``if` `(threeDigit ``%` `8` `=``=` `0``):``        ``count``+``=``1` `    ``# required count``    ``# of rotations``    ``return` `count`  `# Driver Code``if` `__name__``=``=``'__main__'``:``    ``n ``=` `"43262488612"``    ``print``(``"Rotations:"``,countRotationsDivBy8(n))` `# This code is contributed by mits.`

Output:

`Rotations: 4`

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

Please refer complete article on Count rotations divisible by 8 for more details!

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