Enqueue Using Linked List in DSA Python - Time & Space Complexity

Time Complexity: Enqueue Using Linked List

O(1)

Understanding Time Complexity

We want to understand how the time to add an item to a queue using a linked list changes as the queue grows.

How does the number of steps needed to add one item change when the queue gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class Queue:
    def __init__(self):
        self.front = None
        self.rear = None

    def enqueue(self, data):
        new_node = Node(data)
        if self.rear is None:
            self.front = self.rear = new_node
            return
        self.rear.next = new_node
        self.rear = new_node

This code adds a new item to the end of a queue implemented with a linked list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Adding a new node by updating pointers.
• How many times: This happens once per enqueue call, no loops or traversals.

How Execution Grows With Input

Adding one item always takes the same number of steps, no matter how big the queue is.

Input Size (n)	Approx. Operations
10	5-6 steps
100	5-6 steps
1000	5-6 steps

Pattern observation: The time to add one item stays about the same regardless of queue size.

Final Time Complexity

Time Complexity: O(1)

This means adding an item takes a constant amount of time no matter how many items are already in the queue.

Common Mistake

[X] Wrong: "Adding an item takes longer as the queue grows because we have to find the end each time."

[OK] Correct: Because we keep a pointer to the end (rear), we add the new item directly without searching, so time stays constant.

Interview Connect

Knowing that enqueue in a linked list queue is fast and constant time shows you understand efficient data structure design, a key skill in interviews.

Self-Check

"What if we did not keep a rear pointer and had to find the end each time? How would the time complexity change?"

Practice

1. What does the enqueue operation do in a linked list implementation of a queue?

easy

A. Adds a new element at the end (tail) of the linked list

B. Removes the first element from the linked list

C. Adds a new element at the beginning (head) of the linked list

D. Deletes the last element from the linked list

Solution

1. Step 1: Understand enqueue operation in queue

   Enqueue means adding an element to the queue, which is done at the end (tail) to maintain FIFO order.

2. Step 2: Linked list enqueue behavior

   In a linked list, adding at the tail means creating a new node and linking it as the last node.

3. Final Answer:

   Adds a new element at the end (tail) of the linked list -> Option A

4. Quick Check:

   Enqueue = Add at tail [OK]

Hint: Enqueue always adds at the tail in linked list queue [OK]

Common Mistakes:

• Confusing enqueue with dequeue operation
• Adding new element at head instead of tail
• Thinking enqueue removes elements

2. Which of the following is the correct syntax to create a new node and enqueue it at the tail in Python linked list?

easy

A. new_node = Node(data); new_node.next = tail; tail = new_node

B. new_node = Node(data); tail.next = new_node; tail = new_node

C. new_node = Node(data); new_node.next = head; head = new_node

D. new_node = Node(data); head.next = new_node; head = new_node

Solution

1. Step 1: Create new node with data

   We create a new node object with the given data.

2. Step 2: Link new node at tail and update tail pointer

   Set current tail's next to new node, then update tail to new node.

3. Final Answer:

   new_node = Node(data); tail.next = new_node; tail = new_node -> Option B

4. Quick Check:

   New node linked at tail [OK]

Hint: Tail.next points to new node, then tail moves to new node [OK]

Common Mistakes:

• Linking new node to head instead of tail
• Not updating tail pointer after enqueue
• Setting new_node.next to tail (wrong direction)

3. What will be the state of the queue after executing this code?

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

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

    def enqueue(self, data):
        new_node = Node(data)
        if not self.tail:
            self.head = self.tail = new_node
        else:
            self.tail.next = new_node
            self.tail = new_node

q = Queue()
q.enqueue(10)
q.enqueue(20)
q.enqueue(30)
current = q.head
result = []
while current:
    result.append(current.data)
    current = current.next
print(result)

medium

A. []

B. [30, 20, 10]

C. [10, 30, 20]

D. [10, 20, 30]

Solution

1. Step 1: Enqueue 10, 20, 30 in order

   First enqueue sets head and tail to node with 10. Next enqueues add nodes at tail with 20 and 30.

2. Step 2: Traverse from head to tail collecting data

   Traversal collects 10, then 20, then 30 in order.

3. Final Answer:

   [10, 20, 30] -> Option D

4. Quick Check:

   Queue order = insertion order [OK]

Hint: Enqueue adds at tail, so order stays insertion order [OK]

Common Mistakes:

• Assuming enqueue adds at head reversing order
• Not initializing head and tail properly
• Confusing output list order

4. Find the error in this enqueue method for a linked list queue:

def enqueue(self, data):
    new_node = Node(data)
    if self.head is None:
        self.head = new_node
    else:
        self.tail.next = new_node
        self.tail = new_node

medium

A. Does not set new_node.next to None

B. Does not update head when queue is empty

C. Does not update tail when queue is empty

D. Does not link new_node to tail.next

Solution

1. Step 1: Check initialization when queue is empty

   When head is None, head is set to new_node but tail is not updated.

2. Step 2: Tail must also point to new_node when queue is empty

   Without updating tail, tail remains None causing errors on next enqueue.

3. Final Answer:

   Does not update tail when queue is empty -> Option C

4. Quick Check:

   Empty queue needs both head and tail set [OK]

Hint: Set both head and tail on empty queue enqueue [OK]

Common Mistakes:

• Updating only head or only tail on empty queue
• Forgetting to set tail.next on non-empty queue
• Assuming new_node.next needs manual None assignment

5. You want to implement a queue using a linked list that supports enqueue and dequeue in O(1) time. Which of the following changes is necessary to the enqueue method to handle the edge case when the queue is empty?

hard

A. Set both head and tail to the new node when queue is empty

B. Set only tail to the new node when queue is empty

C. Set new node's next to head when queue is empty

D. Set tail.next to head when queue is empty

Solution

1. Step 1: Understand queue empty state

   When queue is empty, both head and tail pointers are None.

2. Step 2: Enqueue first element must update both pointers

   To maintain O(1) enqueue and dequeue, head and tail must both point to the new node.

3. Step 3: This ensures correct linking and prevents errors

   Without setting both, dequeue or next enqueue operations may fail or cause errors.

4. Final Answer:

   Set both head and tail to the new node when queue is empty -> Option A

5. Quick Check:

   Empty queue enqueue sets head and tail [OK]

Hint: Empty queue needs head and tail set to new node [OK]

Common Mistakes:

• Setting only tail or only head on empty enqueue
• Linking new node incorrectly to head or tail
• Not handling empty queue edge case