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.
// 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. |
Output:
Rotations: 2
Time Complexity : O(n) where n is number of digits in input number.
Auxiliary Space: O(1)
Please refer complete article on Count rotations divisible by 4 for more details!