Question

I need a container to store pairs, I have two operations:

1. update value by key
2. get the key with maximum value.

For the first operation, map is a good structure. For the second operation, seems priority queue is a good one. How would you do this? Is there anyway to have both of the two operations done without having a O(n) loop?
Thanks.

Answer 1

An asymptotically efficient solution to this would be to use a combination of a hash table and a Fibonacci heap. You would use the hash table to be able to access the value associated with any particular key in O(1) time, and would use the Fibonacci heap to be able to quickly look up the key/value pair with the lowest value (doing so in O(1)).

If you want to change the value associated with a key, if you are increasing the value, you can do so in (amortized) O(1) time by using the increase-key operation on the Fibonacci heap, which has O(1) amortized time. If you are decreasing the value, it will take O(log n) time to delete the element out of the Fibonacci heap and then reinsert it.

Overall, this gives a time complexity of

• Insert a new element: O(1) for hash, O(1) for insert into Fibonacci heap: O(1) time.
• Delete an element: O(1) for hash, O(log n) for delete from Fibonacci heap: O(log n) time.
• Find top element: O(1) for lookup in Fibonacci heap: O(1) time.
• Increase a value: O(1) for hash, O(1) for increase-key: O(1) time.
• Decrease a value: O(1) for hash, O(log n) for delete/insert: O(log n) time.

Hope this helps!