Time Complexity of Python `rfind()`

Question

I'm wondering if using index or rindex would increase the time complexity to O(n^2) in a loop such as:

for c in s:
    if s.index(c) == s.rindex(c):
        return c

I'm not proficient in C. However, I researched the source code for rindex. Which can be found here . I did notice that they are using some sort of nested while loop, which technically would make the time complexity O(n^2), but that is not always the case.

Edit: Also, the code above seems to be more efficient than something like this:

 h1 = dict()
    
 for index, element in enumerate(s):
     h1[element] = h1.get(element, 0) + 1
    
 for key, value in h1.items():
     if value == 1:
         return key

That said, please enlighten me in understanding if the code above is O(n) or O(n^2) and why it is more efficient.

Tested with a string of length 100k.

I believe the code is O(n^2) due to using an O(n) inside a for a loop. But please correct me if I'm wrong.

Answer 1

I agree, you are preforming an O(n) operation with the for loop, and the s.rindex(c) is preforming also an O(n) operation. Therefore, them multiplied is O(n^2).