Python：计算字典中具有O（logN）中给定值的项数

Question

I want to count the number of items in a dictionary with the given value (suppose the value in dictionary is just number), I've searched online and find two approaches, the first one :

sum(x == chosen_value for x in d.values())

the second approach is using Counter in Collections module.

However, I think the running time of both approaches is O(N) , where N is the total number of items in the dictionary. I want to find out a way to do this in O(logN) , is it possible?

Thanks in advance for any help and suggestion!

Update:

Thanks for all the quick reply! It cannot be done in O(logN) . I may use binary tree to store the (key,value) pairs instead.

Answer 1

No. Why would you expect it to be possible? It might be if you had a binary search tree, but dicts are unordered, so you have to iterate through the values.

Answer 2

You have to read every value in a given dictionary if you have no prior knowledge of these values, which is the normal case. So in the usual case, it has to be O(N) .

Answer 3

You could maintain another dictionary which maps values to count. That would give you the counts that you seek in O(1). The idea would be to implement a higher order data structure that acts like a dictionary, but adds the capability to return counts for values by maintaining another dictionary internally.

I realize that this is probably not what you are asking. Just putting it out there, just on the off chance.