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++

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// 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;
}

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Java

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// 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.

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Python3

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# 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.

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C#

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// 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.

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PHP

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<?php
// PHP program to count all 
// rotations divisible by 8
  
// function to count of all 
// rotations divisible by 8
function countRotationsDivBy8($n)
{
    $len = strlen($n);
    $count = 0;
  
    // For single digit number
    if ($len == 1) 
    {
        $oneDigit = $n[0] - '0';
        if ($oneDigit % 8 == 0)
            return 1;
        return 0;
    }
  
    // For two-digit numbers 
    // (considering all pairs)
    if ($len == 2) 
    {
  
        // first pair
        $first = ($n[0] - '0') * 10 + 
                 ($n[1] - '0');
  
        // second pair
        $second = ($n[1] - '0') * 10 + 
                  ($n[0] - '0');
  
        if ($first % 8 == 0)
            $count++;
        if ($second % 8 == 0)
            $count++;
        return $count;
    }
  
    // considering all 
    // three-digit sequences
    $threeDigit;
    for ($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
$n = "43262488612";
echo "Rotations: "
      countRotationsDivBy8($n);
  
// This code is contributed by mits.
?>

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Output:

Rotations: 4

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



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Improved By : nitin mittal, Mithun Kumar



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