Time Complexity: O(log₁₀N + 1), where N is the input number. The time complexity is determined by the number of digits in the input integer N. In the worst case when N is a multiple of 10 the number of digits in N is log₁₀N + 1.

In the while loop we divide N by 10 until it becomes 0 which takes log₁₀N iterations.
In each iteration of the while loop we perform constant time operations like division and increment the counter.

Space Complexity : O(1), as only a constant amount of additional memory for the counter regardless of size of the input number.

Optimal Approach

Algorithm

The logarithmic base 10 of a positive integers gives the number of digits in n. We add 1 to the result to ensure that the count is correct even for numbers that are powers of 10.

We cast the result to an integer to ensure that any fractional part is discarded giving the exact count of digits.

Algorithm

Initialise a variable to store the count of digits of the number.
The count of digits can be calculated using log₁₀ N + 1.
1. log₁₀ N operation gives the logarithmic base 10 of which returns the power to which 10 must be raised to, to be equal to N.
2. We add 1 to the result which accounts for the possibility that N itself is a power of 10.
3. Financially cast the result to an integer to ensure that it is rounded down to the nearest whole number.
Return the value of count which represents the count of digits in the input integer N

Code