Python3 Program for Count rotations divisible by 8
Last Updated :
09 Jun, 2022
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
def countRotationsDivBy8(n):
l = len (n)
count = 0
if (l = = 1 ):
oneDigit = int (n[ 0 ])
if (oneDigit % 8 = = 0 ):
return 1
return 0
if (l = = 2 ):
first = int (n[ 0 ]) * 10 + int (n[ 1 ])
second = int (n[ 1 ]) * 10 + int (n[ 0 ])
if (first % 8 = = 0 ):
count + = 1
if (second % 8 = = 0 ):
count + = 1
return count
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
threeDigit = ( int (n[l - 1 ]) * 100 +
int (n[ 0 ]) * 10 +
int (n[ 1 ]))
if (threeDigit % 8 = = 0 ):
count + = 1
threeDigit = ( int (n[l - 2 ]) * 100 +
int (n[l - 1 ]) * 10 +
int (n[ 0 ]))
if (threeDigit % 8 = = 0 ):
count + = 1
return count
if __name__ = = '__main__' :
n = "43262488612"
print ( "Rotations:" ,countRotationsDivBy8(n))
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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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