## 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 **Node**s are in the circular linked list. If we have 6,000 **Node**s, 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->...]