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 

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 <bits/stdc++.h>
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[0])
          
        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[0])
    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] - '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] - '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

<?php
// PHP program to count all 
// rotation divisible by 4.
  
// Returns count of all
// rotations divisible
// by 4
function countRotations($n)
{
    $len = strlen($n);
  
    // For single digit number
    if ($len == 1)
    {
        $oneDigit = $n[0] - '0';
  
        if ($oneDigit % 4 == 0)
            return 1;
  
        return 0;
    }
  
    // At-least 2 digit
    // number (considering all
    // pairs)
    $twoDigit;$count = 0;
    for ($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] - '0');
  
    if ($twoDigit % 4 == 0)
        $count++;
  
    return $count;
}
  
// Driver Code
$n = "4834";
echo "Rotations: " ,
      countRotations($n);
  
// This code is contributed by ajit
?>


Output:

Rotations: 2

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

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : nitin mittal, jit_t




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