Enqueue Using Linked List in DSA C# - 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 grow when we add more items?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public class Node {
    public int Data;
    public Node Next;
    public Node(int data) { Data = data; Next = null; }
}

public class Queue {
    private Node front, rear;

    public void Enqueue(int data) {
        Node newNode = new Node(data);
        if (rear == null) {
            front = rear = newNode;
            return;
        }
        rear.Next = newNode;
        rear = newNode;
    }
}
    

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 at the end by updating pointers.
• How many times: This happens once per enqueue call, no loops or recursion inside enqueue.

How Execution Grows With Input

Each enqueue operation takes the same number of steps regardless of queue size.

Input Size (n)	Approx. Operations
10	About 3 pointer updates
100	About 3 pointer updates
1000	About 3 pointer updates

Pattern observation: The work stays the same no matter how big the queue is.

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: "Enqueue takes longer as the queue grows because we have to find the end each time."

[OK] Correct: Because we keep a pointer to the last node (rear), we add 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 only had a front pointer? 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 an element at the front of the linked list

B. Removes an element from the front of the linked list

C. Deletes the last element of the linked list

D. Adds an element at the end of the linked list

Solution

1. Step 1: Understand enqueue operation

   Enqueue means adding an element to the queue.

2. Step 2: Identify linked list behavior for enqueue

   In a linked list queue, enqueue adds the new node at the end (tail).

3. Final Answer:

   Adds an element at the end of the linked list -> Option D

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 element at the front instead of the end
• Thinking enqueue removes elements

2. Which of the following is the correct syntax to create a new node and enqueue it in a linked list queue in C#?

easy

A. Node newNode = new Node(data); tail.next = newNode; tail = newNode;

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

C. Node newNode = new Node(); tail = newNode; tail.next = data;

D. Node newNode = new Node(data); head.next = newNode; head = newNode;

Solution

1. Step 1: Create new node with data

   Use 'new Node(data)' to create a node with the given data.

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

   Set 'tail.next = newNode' and then update 'tail = newNode' to keep tail at last node.

3. Final Answer:

   Node newNode = new Node(data); tail.next = newNode; tail = newNode; -> Option A

4. Quick Check:

   Correct node creation and tail update [OK]

Hint: Use 'new' keyword and update tail pointer [OK]

Common Mistakes:

• Forgetting 'new' keyword when creating node
• Updating head instead of tail
• Assigning data to next pointer

3. What will be the state of the queue after executing the following C# code?

QueueLinkedList queue = new QueueLinkedList();
queue.Enqueue(10);
queue.Enqueue(20);
queue.Enqueue(30);

Assuming the queue prints elements from head to tail.

medium

A. 10 -> 20 -> 30 -> null

B. 30 -> 20 -> 10 -> null

C. null -> 10 -> 20 -> 30

D. 20 -> 30 -> 10 -> null

Solution

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

   Each enqueue adds element at the tail, so order is 10, then 20, then 30.

2. Step 2: Print queue from head to tail

   Head points to 10, followed by 20, then 30, ending with null.

3. Final Answer:

   10 -> 20 -> 30 -> null -> Option A

4. Quick Check:

   Enqueue order preserved from head to tail [OK]

Hint: Enqueue adds at tail, print from head to tail [OK]

Common Mistakes:

• Reversing the order of elements
• Confusing head and tail pointers
• Assuming enqueue adds at head

4. Identify the error in the following enqueue method for a linked list queue in C#:

public void Enqueue(int data) {
    Node newNode = new Node(data);
    if (head == null) {
        head = newNode;
    } else {
        tail.next = newNode;
    }
}

medium

A. New node should be created outside the method

B. The tail pointer is not updated after adding new node

C. The head pointer should be updated instead of tail

D. The if condition should check tail instead of head

Solution

1. Step 1: Check tail update after enqueue

   After adding newNode at tail.next, tail must be updated to newNode.

2. Step 2: Identify missing tail update

   Code does not update tail, so tail still points to old last node, causing errors.

3. Final Answer:

   The tail pointer is not updated after adding new node -> Option B

4. Quick Check:

   Always update tail after enqueue [OK]

Hint: Update tail pointer after adding new node [OK]

Common Mistakes:

• Forgetting to update tail pointer
• Updating head instead of tail
• Creating node outside enqueue method

5. Given a linked list queue with elements 5 -> 15 -> 25 -> null, what will be the queue state after enqueueing 35 and then dequeueing one element?

hard

A. 25 -> 35 -> null

B. 5 -> 15 -> 25 -> 35 -> null

C. 15 -> 25 -> 35 -> null

D. 35 -> 5 -> 15 -> 25 -> null

Solution

1. Step 1: Enqueue 35 at tail

   Queue becomes 5 -> 15 -> 25 -> 35 -> null after enqueue.

2. Step 2: Dequeue removes from head

   Removing 5 from head leaves 15 -> 25 -> 35 -> null.

3. Final Answer:

   15 -> 25 -> 35 -> null -> Option C

4. Quick Check:

   Enqueue adds tail, dequeue removes head [OK]

Hint: Enqueue tail, dequeue head to update queue [OK]

Common Mistakes:

• Removing from tail instead of head
• Not updating head after dequeue
• Adding new element at head instead of tail