Python3 Program to Generate all rotations of a number

Given an integer n, the task is to generate all the left shift numbers possible. A left shift number is a number that is generated when all the digits of the number are shifted one position to the left and the digit at the first position is shifted to the last.
Examples:

Input: n = 123 
Output: 231 312
Input: n = 1445 
Output: 4451 4514 5144 

Approach:

• Assume n = 123.
• Multiply n with 10 i.e. n = n * 10 = 1230.
• Add the first digit to the resultant number i.e. 1230 + 1 = 1231.
• Subtract (first digit) * 10^k from the resultant number where k is the number of digits in the original number (in this case, k = 3).
• 1231 – 1000 = 231 is the left shift number of the original number.

Below is the implementation of the above approach: 

Python3

# Python3 implementation of the approach

# function to return the count of digit of n
def numberofDigits(n):
    cnt = 0
    while n > 0:
        cnt += 1
        n //= 10
    return cnt

# function to print the left shift numbers
def cal(num):
    digit = numberofDigits(num)
    powTen = pow(10, digit - 1)

    for i in range(digit - 1):

        firstDigit = num // powTen

        # formula to calculate left shift
        # from previous number
        left = (num * 10 + firstDigit -
               (firstDigit * powTen * 10))
        print(left, end = " ")

        # Update the original number
        num = left

# Driver code
num = 1445
cal(num)

# This code is contributed
# by Mohit Kumar

Output:

4451 4514 5144

Time Complexity: O(log₁₀(num))
Auxiliary Space: O(1)

Please refer complete article on Generate all rotations of a number for more details!