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Python3 Program to Print all possible rotations of a given Array

Python3 Program to Count rotations divisible by 4

Last Updated : 05 Sep, 2024

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Given a large positive number as string, count all rotations of the given number which are divisible by 4.

Examples:

Input: 8
Output: 1

Input: 20
Output: 1
Rotation: 20 is divisible by 4
          02 is not divisible by 4

Input : 13502
Output : 0
No rotation is divisible by 4

Input : 43292816
Output : 5
5 rotations are : 43292816, 16432928, 81643292
                  92816432, 32928164

For large numbers it is difficult to rotate and divide each number by 4. Therefore, 'divisibility by 4' property is used which says that a number is divisible by 4 if the last 2 digits of the number is divisible by 4. Here we do not actually rotate the number and check last 2 digits for divisibility, instead we count consecutive pairs (in circular way) which are divisible by 4.

Illustration:

Consider a number 928160
Its rotations are 928160, 092816, 609281, 160928, 
    816092, 281609.
Now form pairs from the original number 928160
as mentioned in the approach.
Pairs: (9,2), (2,8), (8,1), (1,6), 
         (6,0), (0,9)
We can observe that the 2-digit number formed by the these 
pairs, i.e., 92, 28, 81, 16, 60, 09, are present in the last
2 digits of some rotation.
Thus, checking divisibility of these pairs gives the required
number of rotations. 

Note: A single digit number can directly
be checked for divisibility.

Below is the implementation of the approach.

Python3

# Python3 program to count # all rotation divisible # by 4.  # Returns count of all # rotations divisible # by 4 def countRotations(n) :      l = len(n)      # For single digit number     if (l == 1) :         oneDigit = (int)(n[0])                  if (oneDigit % 4 == 0) :             return 1         return 0               # At-least 2 digit number     # (considering all pairs)     count = 0     for i in range(0, l - 1) :         twoDigit = (int)(n[i]) * 10 + (int)(n[i + 1])                  if (twoDigit % 4 == 0) :             count = count + 1                  # Considering the number     # formed by the pair of     # last digit and 1st digit     twoDigit = (int)(n[l - 1]) * 10 + (int)(n[0])     if (twoDigit % 4 == 0) :         count = count + 1      return count  # Driver program n = "4834" print("Rotations: " ,     countRotations(n))  # This code is contributed by Nikita tiwari.

Output:

Rotations: 2

Time Complexity : O(n) where n is number of digits in input number.

Auxiliary Space: O(1)

Please refer complete article on Count rotations divisible by 4 for more details!

Python3 Program to Print all possible rotations of a given Array

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Article Tags :

Strings
Mathematical
Python
Python Programs
DSA
divisibility
number-digits
rotation
large-numbers

Practice Tags :

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python
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