Since what () is being called recursively, therefore the final call for what () occurs when n = 1, after which A () is called which takes O(1) amount of time. As the final what () ends, (B (n - (n - 2)) gets called, which is B (2) and takes O(1/2) amount of time. After the end of B (2), what (2) ends and the process repeats until B (1/n) is the final function that is called. Therefore, the time complexity is -
T(n) = O(1) + O(1/2) + O(1/3) ... + O(1/n)
But here's the thing. O(1) represents constant time required to perform a task. This is the smallest amount of time required to perform a certain task by the computer. The time required to perform a task can not get smaller than this. Hence, O(1/2), O(1/3), etc. would be practically impossible on any conventional computer. Therefore these operations would rather take O(1) amount of time each. Thus, the total time complexity of the function becomes -
T(n) = O(1) + O(1) + O(1) + ... up to n
T(n) = n * O(1) = O(n)
Hence, the answer is O(n).