# Count rotations divisible by 8

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

 ` `

Output:

```Rotations: 4
```

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