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:
It is difficult to rotate and divide each number by 8 for large numbers. Therefore, the "divisibility by 8" property says that a number is divisible by 8 if the last 3 digits of the number is divisible by 8. Here we do not rotate the number and check the last 8 digits for divisibility, instead, we count consecutive sequences of 3 digits (in a 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.
// 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";
console.log("Rotations: " + countRotationsDivBy8(n));
Output
Rotations: 4
Complexity Analysis:
- 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!