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 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.
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!