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!
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