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# Count rotations divisible by 8

• Difficulty Level : Easy
• 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, â€˜divisibility by 8â€™ 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] - ``'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] - ``'0'``) * 10 + (n[1] - ``'0'``);`   `        ``// second pair` `        ``int` `second = (n[1] - ``'0'``) * 10 + (n[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[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] - ``'0'``) * 10 + ` `                 ``(n[1] - ``'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] - ``'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] - ``'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] - ``'0'``) * ` `                         ``10 + (n[1] - ``'0'``);` `    `  `            ``// second pair` `            ``int` `second = (n[1] - ``'0'``) * ` `                          ``10 + (n[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[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] - ``'0'``) * 10 + ` `                     ``(n[1] - ``'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] - ``'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.
Auxiliary Space: O(1)

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