Top N percent elements in a queue
You can implement this in O(klogk + n) time complexity.

Maintain an external min-heap while iterating the queue. The min-heap shall have at most k elements at a time.

Insert the first k elements in the min-heap. Thereafter compare the present queue element with the top of the heap. If the top is less than present value then delete the top element from the heap and insert this new present element.

At the end the heap shall contain top k elements in the queue.