in Java, How do I peek() the top k elements in a PriorityQueue ?

Question

I see in the documentation, that PriorityQueue.peek() gives me access to the head of the queue in O(1), but what if I need access to the k top elements in the queue ? I would use poll() k times, but that takes O(log(N)) , is there a way to do it in constant time ?

Answer 1

Nope. If you could do it in constant time, you could do a comparison sort in linear time by heapifying an array and then finding the top N items, where N is all of them.