C++ Program to 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

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 <bits/stdc++.h>
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;
}

Output:

Rotations: 4

Time Complexity: O(n), where n is the number of digits in the input number.
Auxiliary Space: O(1)
Please refer complete article on Count rotations divisible by 8 for more details!

C++ Program to Count rotations divisible by 8

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