Count rotations divisible by 4

• 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 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 4int 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 programint 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

 # Python3 program to count# all rotation divisible# by 4. # Returns count of all# rotations divisible# by 4def 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 programn = "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.



Javascript



Output:

Rotations: 2

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

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