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# Count rotations divisible by 8

• 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 8.

Examples:

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```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. ```

## C++

 `// C++ program to count all rotations divisible``// by 8``#include ``using` `namespace` `std;` `// function to count of all rotations divisible``// by 8``int` `countRotationsDivBy8(string n)``{``    ``int` `len = n.length();``    ``int` `count = 0;` `    ``// For single digit number``    ``if` `(len == 1) {``        ``int` `oneDigit = n - ``'0'``;``        ``if` `(oneDigit % 8 == 0)``            ``return` `1;``        ``return` `0;``    ``}` `    ``// For two-digit numbers (considering all``    ``// pairs)``    ``if` `(len == 2) {` `        ``// first pair``        ``int` `first = (n - ``'0'``) * 10 + (n - ``'0'``);` `        ``// second pair``        ``int` `second = (n - ``'0'``) * 10 + (n - ``'0'``);` `        ``if` `(first % 8 == 0)``            ``count++;``        ``if` `(second % 8 == 0)``            ``count++;``        ``return` `count;``    ``}` `    ``// considering all three-digit sequences``    ``int` `threeDigit;``    ``for` `(``int` `i = 0; i < (len - 2); i++) {``        ``threeDigit = (n[i] - ``'0'``) * 100 +``                     ``(n[i + 1] - ``'0'``) * 10 +``                     ``(n[i + 2] - ``'0'``);``        ``if` `(threeDigit % 8 == 0)``            ``count++;``    ``}` `    ``// Considering the number formed by the``    ``// last digit and the first two digits``    ``threeDigit = (n[len - 1] - ``'0'``) * 100 +``                 ``(n - ``'0'``) * 10 +``                 ``(n - ``'0'``);` `    ``if` `(threeDigit % 8 == 0)``        ``count++;` `    ``// Considering the number formed by the last``    ``// two digits and the first digit``    ``threeDigit = (n[len - 2] - ``'0'``) * 100 +``                 ``(n[len - 1] - ``'0'``) * 10 +``                 ``(n - ``'0'``);``    ``if` `(threeDigit % 8 == 0)``        ``count++;` `    ``// required count of rotations``    ``return` `count;``}` `// Driver program to test above``int` `main()``{``    ``string n = ``"43262488612"``;``    ``cout << ``"Rotations: "``         ``<< countRotationsDivBy8(n);``    ``return` `0;``}`

## Java

 `// Java program to count all``// rotations divisible by 8``import` `java.io.*;` `class` `GFG``{``    ``// function to count of all``    ``// rotations divisible by 8``    ``static` `int` `countRotationsDivBy8(String n)``    ``{``        ``int` `len = n.length();``        ``int` `count = ``0``;``    ` `        ``// For single digit number``        ``if` `(len == ``1``) {``            ``int` `oneDigit = n.charAt(``0``) - ``'0'``;``            ``if` `(oneDigit % ``8` `== ``0``)``                ``return` `1``;``            ``return` `0``;``        ``}``    ` `        ``// For two-digit numbers``        ``// (considering all pairs)``        ``if` `(len == ``2``) {``    ` `            ``// first pair``            ``int` `first = (n.charAt(``0``) - ``'0'``) *``                        ``10` `+ (n.charAt(``1``) - ``'0'``);``    ` `            ``// second pair``            ``int` `second = (n.charAt(``1``) - ``'0'``) *``                         ``10` `+ (n.charAt(``0``) - ``'0'``);``    ` `            ``if` `(first % ``8` `== ``0``)``                ``count++;``            ``if` `(second % ``8` `== ``0``)``                ``count++;``            ``return` `count;``        ``}``    ` `        ``// considering all three-digit sequences``        ``int` `threeDigit;``        ``for` `(``int` `i = ``0``; i < (len - ``2``); i++)``        ``{``            ``threeDigit = (n.charAt(i) - ``'0'``) * ``100` `+``                        ``(n.charAt(i + ``1``) - ``'0'``) * ``10` `+``                        ``(n.charAt(i + ``2``) - ``'0'``);``            ``if` `(threeDigit % ``8` `== ``0``)``                ``count++;``        ``}``    ` `        ``// Considering the number formed by the``        ``// last digit and the first two digits``        ``threeDigit = (n.charAt(len - ``1``) - ``'0'``) * ``100` `+``                    ``(n.charAt(``0``) - ``'0'``) * ``10` `+``                    ``(n.charAt(``1``) - ``'0'``);``    ` `        ``if` `(threeDigit % ``8` `== ``0``)``            ``count++;``    ` `        ``// Considering the number formed by the last``        ``// two digits and the first digit``        ``threeDigit = (n.charAt(len - ``2``) - ``'0'``) * ``100` `+``                    ``(n.charAt(len - ``1``) - ``'0'``) * ``10` `+``                    ``(n.charAt(``0``) - ``'0'``);``        ``if` `(threeDigit % ``8` `== ``0``)``            ``count++;``    ` `        ``// required count of rotations``        ``return` `count;``    ``}``    ` `    ``// Driver program``    ``public` `static` `void` `main (String[] args)``    ``{``        ``String n = ``"43262488612"``;``        ``System.out.println( ``"Rotations: "``                       ``+countRotationsDivBy8(n));``        ` `    ``}``}` `// This code is contributed by vt_m.`

## 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.`

## C#

 `// C# program to count all``// rotations divisible by 8``using` `System;` `class` `GFG {``    ` `    ``// function to count of all``    ``// rotations divisible by 8``    ``static` `int` `countRotationsDivBy8(String n)``    ``{``        ``int` `len = n.Length;``        ``int` `count = 0;``    ` `        ``// For single digit number``        ``if` `(len == 1)``        ``{``            ``int` `oneDigit = n - ``'0'``;``            ``if` `(oneDigit % 8 == 0)``                ``return` `1;``            ``return` `0;``        ``}``    ` `        ``// For two-digit numbers``        ``// (considering all pairs)``        ``if` `(len == 2)``        ``{``    ` `            ``// first pair``            ``int` `first = (n - ``'0'``) *``                         ``10 + (n - ``'0'``);``    ` `            ``// second pair``            ``int` `second = (n - ``'0'``) *``                          ``10 + (n - ``'0'``);``    ` `            ``if` `(first % 8 == 0)``                ``count++;``            ``if` `(second % 8 == 0)``                ``count++;``            ``return` `count;``        ``}``    ` `        ``// considering all three -``        ``// digit sequences``        ``int` `threeDigit;``        ``for` `(``int` `i = 0; i < (len - 2); i++)``        ``{``            ``threeDigit = (n[i] - ``'0'``) * 100 +``                         ``(n[i + 1] - ``'0'``) * 10 +``                         ``(n[i + 2] - ``'0'``);``            ``if` `(threeDigit % 8 == 0)``                ``count++;``        ``}``    ` `        ``// Considering the number formed by the``        ``// last digit and the first two digits``        ``threeDigit = (n[len - 1] - ``'0'``) * 100 +``                     ``(n - ``'0'``) * 10 +``                     ``(n - ``'0'``);``    ` `        ``if` `(threeDigit % 8 == 0)``            ``count++;``    ` `        ``// Considering the number formed``        ``// by the last two digits and``        ``// the first digit``        ``threeDigit = (n[len - 2] - ``'0'``) * 100 +``                     ``(n[len - 1] - ``'0'``) * 10 +``                     ``(n - ``'0'``);``        ``if` `(threeDigit % 8 == 0)``            ``count++;``    ` `        ``// required count of rotations``        ``return` `count;``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main ()``    ``{``        ``String n = ``"43262488612"``;``        ``Console.Write(``"Rotations: "``                      ``+countRotationsDivBy8(n));``        ` `    ``}``}` `// This code is contributed by Nitin Mittal.`

## PHP

 ``

## Javascript

 ``

Output:

`Rotations: 4`

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

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