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Reducing your problem to another one will give you an upper bound on your O(), not a lower one. On the other hand, if you can use any solution to your problem to implement some other algorithm with a well-known lower bound (effectively reducing that problem to yours), that may give you the lower bound you're looking for. <hr> Answer: As others have suggested, you can use the datastructure S to implement sorting: <pre><code>for i in range(input): Insert(S, input[i]) for i in range(input): x = Find_Smallest_Item(S) output[i] = x Delete(S, x) </code></pre> For input of size N this algorithm makes N calls to each of your three operations. However, we know that any general purpose sorting algorithm must have a worst case of O(N log N). This means there will be cases for which the average time of the datastructure calls in the above sort is O(log N) per call. Since that is incompatible with any T() asymptotically better than log N, you have your lower bound. Some notes: <ul> <li>The kind of datastructure described in your problem is called a <a href="http://en.wikipedia.org/wiki/Priority_queue" rel="nofollow">priority queue</a>.</li> <li>Since you're trying to prove a lower bound on any possible priority queue, you can't make an assumption about the implementation. Just because a particular datastructure gives you a particular O() performance, doesn't mean some completely different datastructure couldn't be better.</li> <li>There are many priority queue datastructures that satisfy O(log N) for all calls, so this is actually a tight lower bound.</li> </ul>
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1. POHow to prove a lower bound logn for a data structure?
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1. COMore specifically: consider a way to use this data structure to implement sorting of n items in 3*n data structure operations...
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