Recursive base conversion time complexity analysis - c++

Question

Given an integer p and a destination base b, return a string representation of p in base b. The string should have the least significant bit at the end
^ This is the problem I'm giving myself.
The naive recursive algorithm (in C++) I came up with is as follows:
string convertIntToBaseRecursive(int number, int base) {
// Base case
if (!number) return "";
// Adding least significant digit to "the rest" computed recursively
// Could reverse these operations if we wanted the string backwards.
return convertToBaseRecursive(number / base, base) + to_string(number % base);
}
While the algorithm is incredibly simple, I want to make sure I understand the complexity breakdown. My thoughts are below and I would like to know if they are correct, or wrong, and if they are wrong then knowing where I'm off track would be nice!
Claim:
n = logb(p) is the length of return string
Time complexity: O(n^2)
Space complexity: O(n)
Reasoning:
In order to keep the least significant bit at the end of a string when it is the value we calculate before anything else, we'd either have to:
Compute the string recursively as we are
Keep "shifting" the array every time we calculate a bit so we can add the most recent bit to the front of the string, not the end
Write the string backwards, and reverse it before we return (most efficient)
We're doing the first method in the above C++ algorithm, and the + operator creates a new string at each stack frame. The initial frame creates and returns a string of length n, the next frame creates a string of length n-1, n-2, n-3, and so on. Following this trend (without going into a proof of why 1 + 2 + 3 ... + n = O(n^2), it is clear the time complexity is O(n^2) = O(logb^2(p)). We'll also only need to be storing O(n) things in memory at any time. When the original stack frame resolves (just before algorithm completes) we'll have the memory in terms of a raw string, but before it resolves it will be in terms of a single character (O(1)) + recursive stack frames (O(n)). We do this at each level storing n amounts of single characters until we complete. Therefore the space complexity is O(n).
Of course the more efficient version of this solution would be
string convertIntToBaseIterative(int number, int base) {
string retString = "";
while (number) {
retString += to_string(number % base);
number /= base;
}
// Only needed if least significant
reverse(retString.begin(), retString.end());
return retString;
}
I believe this above solution , where n = logb(p) has:
Time complexity: O(n)
Space complexity: O(n)
Are these analysis correct or am I off somewhere?

Answer 1

Note:
Given the chat room conversation with #user1952500 I had a couple edits to make to his answer based on what we talked about. The following is an edited version of his answer reflecting the latest of what we talked about and what I learned:
Edited answer:
Since the return value has to contain the output, you cannot get a better space complexity than O(n).
Suppose the output string is composed of the following digits in order: a_1, a_2, a_3, ..., a_n. In the
recursive approach (bullet #1), we create a string as follows "a_1" + "a_2" + .... + "a_n", which with
recursion yields O(n^2) time complexity. In bullet #2, the iterative approach constantly pushes characters
to the front of the string like (((...(a_1) + a_2) + a_3) + ... + a_n)))...) which shifts the entire string
on every character addition also yielding O(n^2) time complexity. On your written iterative approach (bullet #3)
the time complexity can be optimized depending on the version of C++ (See notes below).
The string type is not very useful for operations that involve repeated concatenation. In older versions of C++ you
could achieve O(n) time complexity by preallocating a string of size n. In C++11 this
answer indicates that certain append operations can be optimized to be amortized O(1) for a single character. Assuming
this is true the written out iterative version will have O(1) time complexity without any extra work.
Note: To get O(n) time complexity with the recursive version of this algorithm, we could take advantage of the amortized O(1)
character appends and use a single string passed by reference. This would entail the recursive version's function signature
to be re-written as follows:
void convertToBaseRecursive(int number, int base, string& str)

Answer 2

Since the return value has to contain the output, you cannot get a better space complexity than O(n).
Suppose the output string is composed of the following digits in order: a_1, a_2, a_3, ..., a_n. In the recursive approach, we create a string as follows:"a_1" + "a_2" + .... + "a_n". In the iterative approach, we do: (((...(a_1) + a_2) + a_3) + ... + a_n)))...). Hence the time complexity in both cases should be the same at O(n^2) (in C++03. See note below for C++11).
As you see, both approaches have been heavily influenced by implementation details. The string type is not very useful for operations that involve repeated concatenation. If you have a preallocated array of size n, you could get the complexity down to O(n).
Note 1: There is some detail about the append operation. In C++03, the append operation had no specified complexity and could lead to Copy-On_write (if the string could not be extended in place and a relocation was needed). In C++11 CoW and rope-style implementations are disallowed and append should lead to amortized O(1) time per character. Hence in the C++11 case, we should be able to get O(n) time complexity for both implementations.
Note 2: To get a O(n) time complexity with user-defined string implementation (which contains the length), the string needs to be passed by reference in the function. This will lead to the function signature getting changed to:
void convertToBaseRecursive(int number, int base, MyString& str)
This implementation will let the string be shared and updated in-place provided the string uses an array that is pre-allocated.