What does this function return and what are its time/space costs?

def f(nums):
    seen = set()
    for n in nums:
        if n in seen:
            return n
        seen.add(n)
    return -1

First duplicate value; O(n) time, O(n) space

It walks the list once and remembers each number in a set. The first number already in the set is the first repeated value, so it returns it.

One loop, with O(1) set checks inside: O(n) time. The set can grow to n items: O(n) extra memory.

Without the set you would compare every pair: O(n²).

A set turns 'have I seen it?' into O(1).