Why Enqueue Using Linked List in DSA C++?

The Big Idea

What if adding to a line could be done instantly, no matter how long it is?

The Scenario

Imagine you have a line of people waiting to buy tickets, and you try to manage this line by writing down each person's name on a paper every time someone joins. When the line gets long, it becomes hard to keep track and add new names quickly.

The Problem

Writing down each new person manually is slow and mistakes happen easily, like missing a name or writing in the wrong order. It's hard to keep the list updated and organized without a proper system.

The Solution

Using a linked list to enqueue means you add each new person at the end of the line quickly and safely. The linked list keeps track of the last person, so adding a new one is simple and fast without disturbing the rest.

Before vs After

✗ Before

string queue[100];
int rear = -1;
// To add: rear++;
// queue[rear] = newPerson;

✓ After

struct Node {
  string data;
  Node* next;
};
Node* rear = nullptr;
// To add: create new node and link at rear;

What It Enables

This lets you manage growing lines efficiently, adding new items instantly without searching or shifting others.

Real Life Example

Think of a printer queue where documents wait to print. Using a linked list to enqueue new print jobs keeps the queue organized and fast, no matter how many documents are waiting.

Key Takeaways

Manual tracking of queue additions is slow and error-prone.

Linked list enqueue adds new items quickly at the end.

This method keeps queues efficient and easy to manage.

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