Category: Queue

What will we cover in this tutorial?

We will cover what a Queue is. A Queue is a data structure used by computers. It resembles a queue as we know it.

The key things a Queue should have are efficient operations for insertion (enqueue) and removal (dequeue) of the queue. In this tutorial we will understand what a is, how to represent it, and how to implement the efficient operations.

Step 1: Understand a Queue

A Queue is like a queue in real life.

A Queue is used in computer science in many scenarios. One scenario is when a resource is shared among multiple consumers, then a Queue is set in front.

This resembles the real world, where we use queues in the grocery store, pharmacy, you name it. In all stores, where we have more consumers than registers (the resource) to serve.

A Queue serves the principle, First-in-first-out or FIFO.

A simple diagram shows the operations of a Queue.

The above queue shows the direction of the Queue and the order they have been added symbolized with the numbers 1 to 8, where 8 is about to be added.

The operations on the Queue are as follows.

• enqueue Adds an element to the back of the Queue.
• dequeue Removes the element of the front of the Queue.

Normally, a Queue would also have a function is_empty, which checks whether the Queue is empty.

Step 2: How to represent a Queue item with an element

How do you represent the above items of the Queue?

Well, we need to be able to keep an order of the Queue. Let’s try to draw it and see what we can figure out.

Surprised? Well, the Queue goest from left to right, while the arrows between the items go from right to left.

Actually, the items of the Queue can be represented by a simple Node class.

class Node:
    def __init__(self, element=None, next_node=None):
        self.element = element
        self.next_node = next_node

This simple Node class has an element and the pointer to next_node. Hence, the next_node are the arrows in the diagram above. And the element are representing the numbers in the above diagram. Obviously, the element can contain anything.

Step 3: Create a Queue class to represent the enqueue and dequeue operations

Let’s start in the simple.

The above suggest that if we have head and tail pointer, we can have what we need to implement a Queue data structure.

The enqueue function has the special case where the Queue is empty, otherwise it will do as follows.

• enqueue Create a new Node with the element. Point tails next_node at created Node. Then update tail to point at created Node.

Here we can implement it as follows.

class Queue:
    def __init__(self):
        self.head = None
        self.tail = None
    def enqueue(self, element):
        if self.head is None:
            self.head = self.tail = Node(element)
        else:
            node = Node(element)
            self.tail.next_node = node
            self.tail = node

The dequeue is a bit more involved.

• dequeue Get the element from the Node that head points at. If head Node and tail Node is the same (that is if it is the last Node in the Queue), then point tail and head at None. Otherwise, set the head to point at the the next_node head points at.

class Queue:
    def __init__(self):
        self.head = None
        self.tail = None
    def enqueue(self, element):
        if self.head is None:
            self.head = self.tail = Node(element)
        else:
            node = Node(element)
            self.tail.next_node = node
            self.tail = node
    def dequeue(self):
        element = self.head.element
        if self.tail == self.head:
            self.tail = self.head = None
        else:
            self.head = self.head.next_node
        return element

Notice that the dequeue does fail if the Queue is empty. This is dealt with by creating a is_empty function, which returns whether the Queue is empty.

class Queue:
    def __init__(self):
        self.head = None
        self.tail = None
    def enqueue(self, element):
        if self.head is None:
            self.head = self.tail = Node(element)
        else:
            node = Node(element)
            self.tail.next_node = node
            self.tail = node
    def dequeue(self):
        element = self.head.element
        if self.tail == self.head:
            self.tail = self.head = None
        else:
            self.head = self.head.next_node
        return element
    def is_empty(self):
        return self.head is None

That is it.

The full code including a string representation to enable it to be printed

See the full code below. It also contains the __str__(…) function, witch enables it to be printed. Also, how to use it.

class Queue:
    def __init__(self):
        self.head = None
        self.tail = None
    def enqueue(self, element):
        if self.head is None:
            self.head = self.tail = Node(element)
        else:
            node = Node(element)
            self.tail.next_node = node
            self.tail = node
    def dequeue(self):
        element = self.head.element
        if self.tail == self.head:
            self.tail = self.head = None
        else:
            self.head = self.head.next_node
        return element
    def is_empty(self):
        return self.head is None
    def __str__(self):
        if self.head is None:
            return None
        node = self.head
        result = "["
        while node is not None:
            result += str(node.element) + " "
            node = node.next_node
        return result[:-1] + "]"

q = Queue()
for i in range(10):
    q.enqueue(i)
while not q.is_empty():
    print(q.dequeue(), '', end='')

The above code will result in the following output.

[0 1 2 3 4 5 6 7 8 9]
0 1 2 3 4 5 6 7 8 9

What will we cover?

Think about the following challenge. Given a circular linked. Create a queue with constant time operations for enqueue and dequeue. Also, you can only have one pointer to the Circular Linked List.

Consider the above diagram. How will you insert a new element? Okay, that seems easy. But how do you remove the oldest element from it? With constant time!

Yes, that is not that trivial.

Step 1: Create the node class used to represent the circular linked list

There are of course many ways to go about how to create a linked list, and more specifically a circular linked list.

Some implement the classes together, that is both the node and the linked list together. I like to split them from each other. That way, you keep things simple.

Hence, to represent the circular linked list, I suggest the following simple class for a node. Then have the logic of the circular linked list in a different class.

class Node:
    def __init__(self, element=None, next_node=None):
        self.element = element
        self.next_node = next_node

Notice, that the above Node could also be used in a linked list alone. It is a simple building block, which can be used as you please.

Step 2: Create the Queue using a circular linked list

You could create a class for a circular linked list, but we will use it closely from the queue.

The art is to create a queue with only one pointer to the circular linked list structure. The circular linked list can be represented with the Node class above.

The first operation we need is to enqueue an element in the queue. That is to insert an element in the queue.

The special case of inserting the first element, would result in a circular linked list with only one Node, which points at itself.

It is first when we insert the second element it becomes tricky. Which Node should the Queue pointer point at?

This is interesting. We need to be able to insert the element in constant time. Hence, if the queue has more element, it should not need to go all the way around the circular linked list. That makes it necessary, for us to be able to find the last element added with constant number of operations.

Also, the element in the circular linked list point in the direction the were inserted.

This results to that we need to point at the last inserted element.

Assuming in the above that the elements are inserted in the order 1, 2, 3, and 4.

Then we can insert the the elements with the following code.

class Queue:
    def __init__(self):
        self.pointer = None
    def enqueue(self, element):
        if self.pointer is None:
            node = Node(element)
            self.pointer = node
            self.pointer.next_node = node
        else:
            node = Node(element, self.pointer.next_node)
            self.pointer.next_node = node
            self.pointer = node

The code of enqueue shows first the base case, where the circular linked list is empty. Then the case where minimum one Node already exists.

Let’s follow the code of above. Before we call enqueue let’s assume it looks as follows.

First thing is to create a new Node (with the new element, below we use 5) and point it at the queues pointer next_node.

Point the queue pointers next_node at the new Node we created.

Then we update the queue pointer to point at the new Node.

That is it. We still have a circular linked list and only one pointer from the Queue. Notice, that this all happens in constant number of operations. That is, it is independent of how many Nodes are in the circular linked list. If we have 6,000 Nodes, we still only use the 3 operations described above. Hence, enqueue has a O(1) worst case run time.

Next step will see if we can dequeue correctly from the structure, while keeping it intact.

Step 3: How to dequeue from the circular linked list

This seems tricky at first.

But look at the following diagram, where we have 4 elements. They have been inserted by the enqueue function given in Step 2. The first element was 1, then 2, then 3, and finally element 4.

Hence, when we dequeue, we want to return and remove element 1. And the circular linked list should look like this afterwards.

Also, the dequeue operation should be done in constant number of operations.

Obviously, there is a special case, when we only have one element in the queue. Then we remove it and make the queue pointer point at None.

class Queue:
    # Continued...
    def dequeue(self):
        if self.pointer is None:
            return None
        if self.pointer.next_node == self.pointer:
            element = self.pointer.element
            self.pointer = None
            return element
        element = self.pointer.next_node.element
        self.pointer.next_node = self.pointer.next_node.next_node
        return element

Otherwise, we get the element from the queue pointers next_node.

Then we update the queue pointers node to point at next_node.next_node. Notice, that this will keep the structure intact. Also, we keep the dequeue function at a constant number of operations, hence, O(1).

The full code with a print queue function

Here is the full code including a representation to print the queue.

class Node:
    def __init__(self, element=None, next_node=None):
        self.element = element
        self.next_node = next_node

class Queue:
    def __init__(self):
        self.pointer = None
    def enqueue(self, element):
        if self.pointer is None:
            node = Node(element)
            self.pointer = node
            self.pointer.next_node = node
        else:
            node = Node(element, self.pointer.next_node)
            self.pointer.next_node = node
            self.pointer = node
    def dequeue(self):
        if self.pointer is None:
            return None
        if self.pointer.next_node == self.pointer:
            element = self.pointer.element
            self.pointer = None
            return element
        element = self.pointer.next_node.element
        self.pointer.next_node = self.pointer.next_node.next_node
        return element
    def __str__(self):
        if self.pointer is None:
            return "empty"
        first = self.pointer
        node = first.next_node
        result = "["
        while node != first:
            result += str(node.element) + '->'
            node = node.next_node
        return result + str(node.element) + '->...]'

queue = Queue()
for i in range(20):
    queue.enqueue(i)
    print(queue)
for _ in range(21):
    queue.dequeue()
    print(queue)
for i in range(5):
    queue.enqueue(i)
    print(queue)

The beginning of the output will be as follows.

[0->...]
[0->1->...]
[0->1->2->...]
[0->1->2->3->...]
[0->1->2->3->4->...]
[0->1->2->3->4->5->...]