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.
Javascript
<script>// Javascript program to count all// rotations divisible by 8// Function to count of all// rotations divisible by 8function 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 Codelet 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!
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!