Python3 Program to Count rotations divisible by 4

0 0 2 minutes read

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!

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!