Count rotations divisible by 4

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

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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.

C++

 // C++ program to count all rotation divisible // by 4. #include using namespace std;    // Returns count of all rotations divisible // by 4 int countRotations(string n) {     int len = n.length();        // For single digit number     if (len == 1)     {         int oneDigit = n.at(0)-'0';         if (oneDigit%4 == 0)             return 1;         return 0;     }        // At-least 2 digit number (considering all     // pairs)     int twoDigit, count = 0;     for (int i=0; i<(len-1); i++)     {         twoDigit = (n.at(i)-'0')*10 + (n.at(i+1)-'0');         if (twoDigit%4 == 0)             count++;     }        // Considering the number formed by the pair of     // last digit and 1st digit     twoDigit = (n.at(len-1)-'0')*10 + (n.at(0)-'0');     if (twoDigit%4 == 0)         count++;        return count; }    //Driver program int main() {     string n = "4834";     cout << "Rotations: " << countRotations(n) << endl;     return 0; }

Java

 // Java program to count // all rotation divisible // by 4. import java.io.*;    class GFG {            // Returns count of all     // rotations divisible     // by 4     static int countRotations(String n)     {         int len = n.length();                 // For single digit number         if (len == 1)         {           int oneDigit = n.charAt(0)-'0';              if (oneDigit % 4 == 0)               return 1;              return 0;         }                 // At-least 2 digit         // number (considering all         // pairs)         int twoDigit, count = 0;         for (int i = 0; i < (len-1); i++)         {           twoDigit = (n.charAt(i)-'0') * 10 +                      (n.charAt(i+1)-'0');              if (twoDigit%4 == 0)               count++;         }                 // Considering the number         // formed by the pair of         // last digit and 1st digit         twoDigit = (n.charAt(len-1)-'0') * 10 +                    (n.charAt(0)-'0');            if (twoDigit%4 == 0)             count++;                 return count;     }             //Driver program     public static void main(String args[])     {         String n = "4834";         System.out.println("Rotations: " +                           countRotations(n));     } }    // This code is contributed by Nikita tiwari.

Python3

 # Python 3 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)                    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)     if (twoDigit % 4 == 0) :         count = count + 1        return count    # Driver program n = "4834" print("Rotations: " ,     countRotations(n))    # This code is contributed by Nikita tiwari.

C#

 // C# program to count all rotation // divisible by 4. using System;    class GFG {            // Returns count of all     // rotations divisible     // by 4     static int countRotations(String n)     {         int len = n.Length;                // For single digit number         if (len == 1)         {             int oneDigit = n - '0';                    if (oneDigit % 4 == 0)                 return 1;                    return 0;         }                // At-least 2 digit         // number (considering all         // pairs)         int twoDigit, count = 0;         for (int i = 0; i < (len - 1); i++)         {             twoDigit = (n[i] - '0') * 10 +                           (n[i + 1] - '0');                    if (twoDigit % 4 == 0)                 count++;         }                // Considering the number         // formed by the pair of         // last digit and 1st digit         twoDigit = (n[len - 1] - '0') * 10 +                                (n - '0');            if (twoDigit % 4 == 0)             count++;                return count;     }            //Driver program     public static void Main()     {         String n = "4834";         Console.Write("Rotations: " +                     countRotations(n));     } }    // This code is contributed by nitin mittal.

PHP



Output:

Rotations: 2

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

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Improved By : nitin mittal, jit_t

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