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
// Java program to count all // rotations divisible by 8import 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!
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