Java Program for 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.
Java
import java.io.*;
class GFG
{
static int countRotationsDivBy8(String n)
{
int len = n.length();
int count = 0 ;
if (len == 1 ) {
int oneDigit = n.charAt( 0 ) - '0' ;
if (oneDigit % 8 == 0 )
return 1 ;
return 0 ;
}
if (len == 2 ) {
int first = (n.charAt( 0 ) - '0' ) *
10 + (n.charAt( 1 ) - '0' );
int second = (n.charAt( 1 ) - '0' ) *
10 + (n.charAt( 0 ) - '0' );
if (first % 8 == 0 )
count++;
if (second % 8 == 0 )
count++;
return count;
}
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++;
}
threeDigit = (n.charAt(len - 1 ) - '0' ) * 100 +
(n.charAt( 0 ) - '0' ) * 10 +
(n.charAt( 1 ) - '0' );
if (threeDigit % 8 == 0 )
count++;
threeDigit = (n.charAt(len - 2 ) - '0' ) * 100 +
(n.charAt(len - 1 ) - '0' ) * 10 +
(n.charAt( 0 ) - '0' );
if (threeDigit % 8 == 0 )
count++;
return count;
}
public static void main (String[] args)
{
String n = "43262488612" ;
System.out.println( "Rotations: "
+countRotationsDivBy8(n));
}
}
|
Output:
Rotations: 4
Time Complexity : O(n), where n is the number of digits in input number.
Auxiliary Space: O(1)
Please refer complete article on Count rotations divisible by 8 for more details!
Last Updated :
09 Jun, 2022
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...