Lower bounds: queues that return their min elements in $O(1)$ time

Problem Detail: 

First, consider this simple problem --- design a data structure of comparable elements that behaves just like a stack (in particular, push(), pop() and top() take constant time), but can also return its min value in $O(1)$ time, without removing it from the stack. This is easy by maintaining a second stack of min values.

Now, consider the same problem, where the stack is replaced by a queue. This seems impossible because one would need to keep track of $\Theta(n^2)$ values (min values between elements $i$ and $j$ in the queue). True or false ?

Update: $O(1)$ amortized time is quite straightforward as explained in one of the answers (using two min-stacks). A colleague pointed out to me that one can de-amortize such data structures by performing maintenance operations proactively. This is a little tricky, but seems to work.

Asked By : Igor Markov

Answered By : Vicky Chijwani

Yes, it can be done in amortized O(1) time (enqueue, dequeue, minimum). Have a look at the code and explanation given here.

Here are the most relevant parts quoted from the above link (emphasis mine; and of course, all credits for the solution go to the original author, Keith Schwarz):

A FIFO queue class that supports amortized O(1) enqueue, dequeue, and find- min. Here, the find-min returns (but does not remove) the minimum value in the queue. This is not the same as a priority queue, which always removes the smallest element on each dequeue. The min-queue simply allows for constant-time access to that element.

The construction that enables the min-queue to work is based on a classic construction for creating a queue out of two stacks. ... we apply this construction to two min-stacks instead of two regular stacks. The minimum element of the queue can then be found by looking at the minimum element of the two stacks taken together.

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Question Source : http://cs.stackexchange.com/questions/6146