## 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->...]
```

## How to profile a program in Python

In this video we will see how cProfile (default Python library) can help you to get run-times from your Python program.

## Queue vs Python lists

In this video we will compare the performance of a simple Queue implemented directly into Python (no optimisations) with the default Python list.

Can it compare with it on performance?

This is where time complexity analysis come into the picture. A Queue insert and deletion is O(1) time complexity. A Python list used as a queue has O(n) time complexity.

But does the performance and run-time show the same? Here we compare the run-time by using cProfile in Python.

Check out my Course on Linked Lists, Stacks and Queues

## How to Implement a Queue in Python and Compare Performance with a Python list

• What is a Queue?
• Implement a Queue in Python
• Make performance testing of it
• Compare it with performance of a Python list

## What is a Queue?

We all know what a queue is. You go to the grocery store and get spinach, strawberry and bananas for your shake. Then you see a long line of people in front of the register. That line is a queue.

The same holds in programming. You create queues to process data or input of any kind.

## How to implement a Queue in Python

It is easier than you think.

First you create a Node class to represent each node in a queue. A node is an abstraction to represent a point to the next node and the actual element.

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

```

Then you create the class for the Queue.

```class Queue:
def __init__(self):
self.tail = None

def enqueue(self, element):
else:
n = Node(element, self.tail)
self.tail.next_node = n
self.tail = n

def dequeue(self):
else:
return element

def is_empty(self):
```

How does it work. Let’s make a simple example.

```q = Queue()
for i in range(10):
q.enqueue(i)

while not q.is_empty():
print(q.dequeue())
```

Which will output.

```0
1
2
3
4
5
6
7
8
9
```

Yes! You guessed it.

## How do we test performance?

I like to use the cProfile library. It is easy to use and gives informative results.

So how do you test performance? You simply import the cProfile library and use the cProfile.run(…) call.

You also need to do some operations to see how your Queue performs. See the code as an example.

```import cProfile

def profile_queue(n):
q = Queue()
for i in range(n):
q.enqueue(i)
while not q.is_empty():
q.dequeue()

def profile(n):
profile_queue(n)

cProfile.run("profile(100000)")

```

Which will result in the following output.

```   Ordered by: standard name

ncalls  tottime  percall  cumtime  percall filename:lineno(function)
1    0.000    0.000    0.819    0.819 <string>:1(<module>)
100000    0.310    0.000    0.351    0.000 Queue.py:11(enqueue)
100000    0.308    0.000    0.308    0.000 Queue.py:19(dequeue)
100000    0.041    0.000    0.041    0.000 Queue.py:2(__init__)
100001    0.021    0.000    0.021    0.000 Queue.py:27(is_empty)
1    0.132    0.132    0.819    0.819 Queue.py:34(profile_queue)
1    0.000    0.000    0.819    0.819 Queue.py:42(profile)
1    0.000    0.000    0.000    0.000 Queue.py:7(__init__)
1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
1    0.008    0.008    0.008    0.008 {range}

```

The interesting line is line 9, which tells us how much time is spend in the call to profile_queue.

But is the result good?

We need to compare it to other implementations.

## Performance testing the Queue with a Python list

Python lists are used for anything. Can we use a Python list as a Queue. Of course. Let’s try to implement that and compare it to our Queue.

```import cProfile

def profile_queue(n):
q = Queue()
for i in range(n):
q.enqueue(i)
while not q.is_empty():
q.dequeue()

def profile_list_as_queue(n):
q = []
for i in range(n):
q.insert(0,i)
while len(q) > 0:
q.pop()

def profile(n):
profile_queue(n)
profile_list_as_queue(n)

cProfile.run("profile(100000)")
```

How does that compare? Let’s see.

```   Ordered by: standard name

ncalls  tottime  percall  cumtime  percall filename:lineno(function)
1    0.000    0.000    3.680    3.680 <string>:1(<module>)
100000    0.295    0.000    0.331    0.000 Queue.py:11(enqueue)
100000    0.298    0.000    0.298    0.000 Queue.py:19(dequeue)
100000    0.036    0.000    0.036    0.000 Queue.py:2(__init__)
100001    0.019    0.000    0.019    0.000 Queue.py:27(is_empty)
1    0.104    0.104    0.756    0.756 Queue.py:34(profile_queue)
1    0.101    0.101    2.924    2.924 Queue.py:42(profile_list_as_queue)
1    0.000    0.000    3.680    3.680 Queue.py:50(profile)
1    0.000    0.000    0.000    0.000 Queue.py:7(__init__)
100001    0.005    0.000    0.005    0.000 {len}
1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
100000    2.805    0.000    2.805    0.000 {method 'insert' of 'list' objects}
100000    0.012    0.000    0.012    0.000 {method 'pop' of 'list' objects}
2    0.004    0.002    0.004    0.002 {range}

```

Wow. Our Queue is way faster than the Python list.

But how is it comparing in general? Comparing the performance of the Queue and a Python list as a Queue.

While it is difficult to see, the performance of the Queue is O(n) (linear) while the performance of the Python list as a Queue is O(n^2).

Hence, the Queue will outperform the Python list for this use case.