I ran into an interview question recently. no additional info is given into question (maybe default implementation should be used...)
n arbitrary sequences of insert and remove operations on empty min heap (location for delete element is known) has amortized cost of:
A) insert O(1), remove O(log n)
B) insert O(log n), remove O(1)
The option (B) is correct.
I'm surprized when see answer sheet. i know this is tricky, maybe empty heap, maybe knowing location of elements for delete,... i dont know why (A) is false? Why (B) is true?
Because the heap is initially empty, you can't have more deletes than inserts.
An amortized cost of O(1) per deletion and O(log N) per insertion is exactly the same as an amortized cost of O(log N) for both inserts and deletes, because you can just count the deletion cost when you do the corresponding insert.
It does not work the other way around. Since you can have more inserts than deletes, there might not be enough deletes to pay the cost of each insert.