Can someone help me understand the time and space complexity? Can anyone help me understand the time and space complexity here? I know zip() is O(1) and sorted() is O(n). The loop has the operation min() inside but is operating on just 2 values so is that also O(1)? If so then every operation in the loop is O(1) so the overall time complexity is O(n) correct? For space, z could grow to n as a and b grow to n as well so space is O(n)?
a = [3, 1, 2]
b = [1, 2, 3]
c = 4
def foo(a: list[int], b: list[int], c: int) -&gt; int:
    ans = 0
    if not a:
        return ans
    z = sorted(zip(a, b), reverse=True)
    for x, y in z:
        if c &gt; 0:
            x = min(x, c)
            ans += x * y
            c -= x
    return ans
Assuming that sorted() really takes time $O(n)$, then the time and space complexities are indeed $O(n)$. Hoewever there are some catches.
You say that the zip function takes $O(1)$ time. This is technically true, but you should be aware that zip is not creating a list that contains pairs of elements from a and b but just returning an iterator to these pairs. Going through the list implicitly returned by this iterator takes $O(n)$ time, even just to construct it (e.g., list(zip(a,b))). In your particular case this doesn't really matter since the time complexity will be $\Omega(n)$ anyway.
You say that sorted() takes $O(n)$ time. This is not impossible, but it's at least suspicious (have you checked in the python documentation/implementation?). There is no algorithm that can sort a list of $n$ elements in time $O(n)$ by only performing comparisons. The implementation of sorted() would need to be particularly clever and recognize that you are sorting small integers.
Probably sorted() just takes $O(n \log n)$ time and $O(n)$ space, which would result in the same overall time and space complexities for your algorithm.