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.
<script> // Javascript program to count all // rotations divisible by 8 // Function to count of all // rotations divisible by 8 function countRotationsDivBy8(n)
{ let len = n.length;
let count = 0;
// For single digit number
if (len == 1)
{
let oneDigit = n[0] - '0' ;
if (oneDigit % 8 == 0)
return 1;
return 0;
}
// For two-digit numbers
// (considering all pairs)
if (len == 2)
{
// first pair
let first = (n[0] - '0' ) * 10 +
(n[1] - '0' );
// second pair
let second = (n[1] - '0' ) * 10 +
(n[0] - '0' );
if (first % 8 == 0)
count++;
if (second % 8 == 0)
count++;
return count;
}
// Considering all
// three-digit sequences
let threeDigit;
for (let i = 0; i < (len - 2); i++)
{
threeDigit = (n[i] - '0' ) * 100 +
(n[i + 1] - '0' ) * 10 +
(n[i + 2] - '0' );
if (threeDigit % 8 == 0)
count++;
}
// Considering the number
// formed by the last digit
// and the first two digits
threeDigit = (n[len - 1] - '0' ) * 100 +
(n[0] - '0' ) * 10 +
(n[1] - '0' );
if (threeDigit % 8 == 0)
count++;
// Considering the number
// formed by the last two
// digits and the first digit
threeDigit = (n[len - 2] - '0' ) * 100 +
(n[len - 1] - '0' ) * 10 +
(n[0] - '0' );
if (threeDigit % 8 == 0)
count++;
// Required count
// of rotations
return count;
} // Driver Code let n = "43262488612" ;
document.write( "Rotations: " +
countRotationsDivBy8(n));
// This code is contributed by _saurabh_jaiswal </script> |
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!