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Count rotations divisible by 4

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  • Difficulty Level : Easy
  • Last Updated : 08 Jul, 2022
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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




# Python3 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
?>

Javascript




<script>
 
// Javascript program to count all
// rotation divisible by 4.
 
// Returns count of all
// rotations divisible
// by 4
function countRotations(n)
{
    let len = n.length;
 
    // For single digit number
    if (len == 1)
    {
        let oneDigit = n[0] - '0';
 
        if (oneDigit % 4 == 0)
            return 1;
 
        return 0;
    }
 
    // At-least 2 digit
    // number (considering all
    // pairs)
    let twoDigit;
    let count = 0;
     
    for(let 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
let n = "4834";
 
document.write("Rotations: " +
               countRotations(n));
 
// This code is contributed by _saurabh_jaiswal
     
</script>

Output

Rotations: 2

Time Complexity : O(n) where n is number of digits in input number.
Auxiliary Space: O(1)

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


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