Implement a topK function for spenders using a min-heap.
Use the signature def topK(spenders, k):. spenders is a list of [name, amount] pairs with unique names. Return the names of the top k spenders, ordered by amount descending; break equal amounts by name descending. Return an empty list when k is zero. The solution should maintain at most k entries in the heap.
def topK(spenders, k):