7 Priority queues

So far we have seen two ADTs that represent a collection of objects, and support adding and removing objects:

Stacks, where the object removed is always the one most recently inserted (LIFO).
Queues, where the object removed is always the one first inserted (FIFO).

There are many situations, both in real life and in computing applications, where we wish to choose the next “most important” from a collection of people, tasks, or objects. For example, doctors in a hospital emergency room often choose to see next the “most critical” patient rather than the one who arrived first. When scheduling programs for execution in a multitasking operating system, at any given moment there might be several programs (usually called jobs) ready to run. The next job selected is the one with the highest priority. Priority is indicated by a particular value associated with the job (and might change while the job remains in the wait list).

When a collection of objects is organized by importance or priority, we call this a priority queue. A priority queue supports the following operations:

adding a new object to the priority queue
removing the smallest object from the priority queue.

In this chapter, we will see how to implement a priority queue so that both adding and removing the minimum take $O(\log n)$ time.

interface PriorityQueue extends Collection:
    add(x)       // Adds x to the priority queue.
    removeMin()  // Removes and returns the minimum element.
    getMin()     // Returns the minimum element, without removing it.

Note that this API assumes that the priority queue orders the elements in ascending order. There is also the possibility of ordering in descending order – that kind of queue is called a maximum priority queue. If you have a minimum priority queue, it’s straightforward to turn it into a maximum priority queue.

Now let’s look at a couple of applications of priority queues.

Example: Sorting

We can use a priority queue to make an efficient sorting algorithm. To sort a list of items:

First create an empty priority queue, and add all the items to it.
Then repeatedly find and remove the smallest item. The items will come out in ascending order.

Here is an implementation of this algorithm in code:

function pqSort(array):
    pq = new PriorityQueue()
    for each item in array:
        pq.add(item)
    for i = 0 to array.size()-1:
        array[i] = pq.removeMin()

What is the time complexity of this algorithm? Well, for an input list of size $n$ , the algorithm calls add $n$ times and removeMin $n$ times. In a binary heap, add and removeMin both take $O(\log n)$ time. Therefore, the total runtime is $O(n \log n)$ – as efficient as any of the sorting algorithms we have seen so far!

Example: Finding the top 100 items

Suppose that we are running a bank. Every day, every transaction that occurs at the bank is recorded in a list. When the bank closes at the end of the day, we would like to find the 100 highest-valued transactions from that day. How can we do it?

One way is to use sorting. If we store the transactions in an array and sort it by value, then the highest-value transactions will be at the end of the array. If there are n transactions in total, then transactions number $n-100\ldots n-1$ are the ones we need. If we use an efficient sorting algorithm, this will take $O(n \log n)$ time. (More generally, this gives us an algorithm for finding the largest $k$ items in a list of $n$ items, in $O(n \log n)$ time.)

Now suppose that we want to monitor the transactions throughout the day. At any point, we want to be able to find the 100 highest-valued transactions so far today. How can we do this?

We could still use the sorting approach, but we would need to sort the list of transactions every time we wanted to find the 100 top transactions. This may be prohibitively expensive if there are a lot of transactions: it takes $O(n \log n)$ time every time we do it.

We can do better with the help of a priority queue. The idea is to have a priority queue that holds the 100 highest-value transactions only. Whenever a new transaction comes in, we need to update the priority queue accordingly:

If the priority queue has fewer than 100 transactions (i.e. there have been fewer than 100 transactions so far today), then add the new transaction to the priority queue.
Otherwise, if the new transaction is greater in value than the lowest-valued of the top 100 transactions, then remove that transaction and add the new transaction.
Otherwise, don’t add the new transaction to the priority queue (it’s not in the top 100).

Notice that in step 2, we are comparing the new transaction to the lowest-valued of the top 100 transactions – if the transactions are ordered by value, then this transaction can be found by calling getMin, and removed using removeMin. So this algorithm can be implemented efficiently using a priority queue.

In fact, we can simplify these three steps into two steps. First, we add the new transaction to the priority queue. This might make the priority queue grow to 101 transactions. If so, we remove the lowest-valued transaction. Here it is in code:

class Top100Transactions:
    // Assume that the Transaction class implements comparisons
    // by comparing the value of the transaction.
    Top100Transactions():
        this.pq = new PriorityQueue()

    // Add a new transaction to the priority queue.
    add(transaction):
        pq.add(transaction)
        // If the priority queue grows to 101 transactions,
        // cut it down to 100 by removing the smallest-valued one.
        if pq.size() > 100:
            pq.removeMin()

    // Return the top 100 transactions.
    top100():
        return pq.iterator()

What is the complexity of add? Well, in fact it takes constant time, because the priority queue has a constant maximum size of 100 elements. If we generalize this problem to keeping track of the top $k$ transactions, then the complexity of add is $O(\log k)$ .