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.
Python3
# Python3 program to count all # rotations divisible by 8# function to count of all # rotations divisible by 8def countRotationsDivBy8(n): l = len(n) count = 0 # For single digit number if (l == 1): oneDigit = int(n[0]) if (oneDigit % 8 == 0): return 1 return 0 # For two-digit numbers # (considering all pairs) if (l == 2): # first pair first = int(n[0]) * 10 + int(n[1]) # second pair second = int(n[1]) * 10 + int(n[0]) if (first % 8 == 0): count+=1 if (second % 8 == 0): count+=1 return count # considering all # three-digit sequences threeDigit=0 for i in range(0,(l - 2)): threeDigit = (int(n[i]) * 100 + int(n[i + 1]) * 10 + int(n[i + 2])) if (threeDigit % 8 == 0): count+=1 # Considering the number # formed by the last digit # and the first two digits threeDigit = (int(n[l - 1]) * 100 + int(n[0]) * 10 + int(n[1])) if (threeDigit % 8 == 0): count+=1 # Considering the number # formed by the last two # digits and the first digit threeDigit = (int(n[l - 2]) * 100 + int(n[l - 1]) * 10 + int(n[0])) if (threeDigit % 8 == 0): count+=1 # required count # of rotations return count# Driver Codeif __name__=='__main__': n = "43262488612" print("Rotations:",countRotationsDivBy8(n))# This code is contributed by mits. |
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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