Given a number n, the task is to count all rotations of the given number which are odd and even.
Input: n = 1234 Output: Odd = 2, Even = 2 Total rotations: 1234, 2341, 3412, 4123 Odd rotations: 2341 and 4123 Even rotations: 1234 and 3412 Input: n = 246 Output: Odd = 0, Even = 3
Efficient Approach: For large numbers, it is difficult to rotate and check whether it is odd or not for every rotation. Hence, in this approach, check the count of odd digits and even digits present in the number. These will be the answer to this problem.
Below is the implementation of the above approach:
Odd = 2 Even = 2
Time Complexity: O(n)
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