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 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!