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.

struct Node {
    int data;
    Node* next;
};

void enqueue(Node*& front, Node*& rear, int value) {
    Node* newNode = new Node{value, nullptr};
    if (rear == nullptr) {
        front = rear = newNode;
    } else {
        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: Creating a new node and updating pointers.
• How many times: Each enqueue call does this once, no loops or recursion involved.

How Execution Grows With Input

Adding one item always takes the same number of steps, no matter how many items are already in the queue.

Input Size (n)	Approx. Operations
10	1 (create node + update pointers)
100	1 (same steps as for 10)
1000	1 (still just one set of steps)

Pattern observation: The time to add one item stays 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 big the queue is.

Common Mistake

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

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

Interview Connect

Knowing that enqueue in a linked list queue is 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 an element at the end (rear) of the linked list

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

C. Deletes the entire linked list

D. Adds an element at the front (head) of the linked list

Solution

1. Step 1: Understand enqueue operation in queue

   Enqueue means adding an element to the queue.

2. Step 2: Identify linked list position for enqueue

   In a linked list queue, enqueue adds at the rear (end) to maintain FIFO order.

3. Final Answer:

   Adds an element at the end (rear) of the linked list -> Option A

4. Quick Check:

   Enqueue = Add at rear [OK]

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

Common Mistakes:

• Confusing enqueue with dequeue operation
• Adding element at front instead of rear
• 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* temp = Node(data); rear->next = temp; rear = temp;

B. Node temp = new Node(data); rear->next = temp; rear = temp;

C. Node* temp = new Node(data); rear->next = temp; rear = temp;

D. Node* temp = new Node; rear->next = temp; rear = temp;

Solution

1. Step 1: Check node creation syntax in C++

   Correct way to create a node dynamically is Node* temp = new Node(data);.

2. Step 2: Verify linking and updating rear pointer

   Link new node by rear->next = temp; and update rear to new node rear = temp;.

3. Final Answer:

   Node* temp = new Node(data); rear->next = temp; rear = temp; -> Option C

4. Quick Check:

   Use new with pointer and update rear [OK]

Hint: Use 'new' keyword and pointer for node creation [OK]

Common Mistakes:

• Not using pointer (*) for node
• Missing 'new' keyword for dynamic allocation
• Incorrect assignment without pointers

3. What will be the state of the queue after executing the following enqueue operations on an empty linked list queue?

enqueue(10); enqueue(20); enqueue(30);

medium

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

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

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

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

Solution

1. Step 1: Enqueue 10 on empty queue

   Queue becomes 10 -> null.

2. Step 2: Enqueue 20 and 30 sequentially

   Adding 20 at rear: 10 -> 20 -> null; then 30 at rear: 10 -> 20 -> 30 -> null.

3. Final Answer:

   10 -> 20 -> 30 -> null -> Option B

4. Quick Check:

   FIFO order maintained: 10, 20, 30 [OK]

Hint: Enqueue adds at rear, order stays same [OK]

Common Mistakes:

• Reversing order of elements
• Confusing enqueue with push front
• Ignoring null termination

4. Identify the error in the following enqueue function for a linked list queue:

void enqueue(int data) {
  Node* temp = new Node(data);
  if (rear == nullptr) {
    front = temp;
  } else {
    rear->next = temp;
  }
  // Missing update of rear pointer here
}

medium

A. rear pointer is not updated to the new node

B. front pointer should be updated to temp always

C. Node creation syntax is incorrect

D. next pointer of temp should be set to front

Solution

1. Step 1: Check rear pointer update after enqueue

   rear pointer must point to the newly added node to keep track of queue end.

2. Step 2: Identify missing update in code

   Code misses rear = temp; after linking new node.

3. Final Answer:

   rear pointer is not updated to the new node -> Option A

4. Quick Check:

   rear must update to new node [OK]

Hint: Always update rear to new node after enqueue [OK]

Common Mistakes:

• Forgetting to update rear pointer
• Updating front pointer incorrectly
• Setting next pointer wrongly

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

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

C. 25 -> 35 -> null

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

Solution

1. Step 1: Enqueue 35 at rear

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

2. Step 2: Dequeue removes front element 5

   After dequeue, front moves to 15, so queue is 15 -> 25 -> 35 -> null.

3. Final Answer:

   15 -> 25 -> 35 -> null -> Option D

4. Quick Check:

   Enqueue rear, dequeue front; update pointers accordingly [OK]

Hint: Enqueue rear, dequeue front; update pointers accordingly [OK]

Common Mistakes:

• Removing wrong element on dequeue
• Not updating front pointer after dequeue
• Adding new element at front instead of rear