Two sum in one pass

Given a list of prices and a target, find two different positions whose prices add up to the target. Checking every pair works. There is also a way that walks the list only once: for each price ask "have I already seen the price that would complete this pair?" A dictionary from price to position answers that directly. If the answer is no, record the current price and move on.

Write two_sum(prices, target) that returns the two positions as a tuple (earlier, later), or None if no pair works. Exactly one pair works when one exists.

two_sum([2, 7, 11, 15], 9)  ->  (0, 1)
two_sum([1, 2], 10)         ->  None

def two_sum(prices, target):
    ...