Dequeue Using Linked List in DSA C++ - Build from Scratch

Dequeue Using Linked List

📖 Scenario: You are building a simple system to manage a line of customers where people can join or leave from both ends. This is like a double-ended queue (deque) using a linked list.

🎯 Goal: Create a deque using a linked list in C++. You will add nodes at the front and rear, remove nodes from front and rear, and finally print the current state of the deque.

📋 What You'll Learn

Create a linked list node structure with integer data and pointers to next and previous nodes.

Implement functions to insert at front and rear of the deque.

Implement functions to delete from front and rear of the deque.

Print the deque from front to rear showing all elements.

💡 Why This Matters

🌍 Real World

Deque structures are used in real-world scenarios like task scheduling, undo operations in software, and managing lines where people can join or leave from both ends.

💼 Career

Understanding linked lists and deque operations is important for software developers working on system design, performance optimization, and implementing custom data structures.

1

Create Node Structure and Initialize Dequeue

Create a struct called Node with an int data, a Node* next, and a Node* prev. Then create two Node* pointers called front and rear and initialize both to nullptr.

DSA C++

// Define Node struct and initialize front and rear pointers
// Your code here

Hint

Think of a node as a box holding a number and links to the next and previous boxes.

2

Add Helper Function to Create New Node

Write a function called createNode that takes an int value and returns a Node*. Inside, create a new Node, set its data to value, and set next and prev to nullptr. Return the new node.

DSA C++

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

Node* front = nullptr;
Node* rear = nullptr;

// Write createNode function below
// Your code here

Hint

This function helps you make new boxes with the number inside and no links yet.

3

Implement Insert at Front and Rear

Write two functions: insertFront and insertRear. Each takes an int value. For insertFront, create a new node and add it at the front of the deque, updating front and rear pointers correctly. For insertRear, create a new node and add it at the rear, updating pointers similarly.

DSA C++

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

Node* front = nullptr;
Node* rear = nullptr;

Node* createNode(int value) {
    Node* newNode = new Node;
    newNode->data = value;
    newNode->next = nullptr;
    newNode->prev = nullptr;
    return newNode;
}

// Write insertFront and insertRear functions below
// Your code here

Hint

Remember to update both front and rear pointers when the deque is empty.

4

Print Dequeue from Front to Rear

Write a function called printDequeue that prints all elements from front to rear separated by -> and ending with null. Then call insertFront(10), insertRear(20), insertFront(5), and finally call printDequeue() to show the deque.

DSA C++

Hint

Traverse from front to rear and print each node's data followed by ' -> ', then print 'null' at the end.

Practice

1. What is the main advantage of implementing a Dequeue using a linked list over an array?

easy

A. It allows insertion and deletion from both ends in O(1) time without shifting elements.

B. It uses less memory than an array.

C. It automatically sorts the elements.

D. It restricts insertion to only one end.

Solution

1. Step 1: Understand Dequeue operations

   A dequeue allows insertion and deletion from both front and rear ends.

2. Step 2: Compare linked list and array

   Linked list supports O(1) insertion/deletion at both ends without shifting, unlike arrays.

3. Final Answer:

   It allows insertion and deletion from both ends in O(1) time without shifting elements. -> Option A

4. Quick Check:

   Linked list dequeue = O(1) insert/delete both ends [OK]

Hint: Linked list avoids shifting, so both ends are fast [OK]

Common Mistakes:

• Thinking arrays allow O(1) insert/delete at both ends
• Assuming linked list sorts elements automatically
• Confusing memory usage advantages

2. Which of the following is the correct syntax to insert an element at the front of a dequeue implemented using a linked list in C++?

easy

A. void insertFront(int data) { front = data; }

B. void insertFront(int data) { Node* newNode = new Node(data); newNode->next = front; front = newNode; }

C. void insertFront(int data) { front->data = data; }

D. void insertFront(int data) { delete front; front = new Node(data); }

Solution

1. Step 1: Create new node with data

   We allocate a new node and assign the data to it.

2. Step 2: Link new node to current front and update front pointer

   Set newNode->next to current front, then update front to newNode.

3. Final Answer:

   void insertFront(int data) { Node* newNode = new Node(data); newNode->next = front; front = newNode; } -> Option B

4. Quick Check:

   New node points to old front, front updated [OK]

Hint: New node next points to old front, then front updated [OK]

Common Mistakes:

• Assigning data directly to front pointer
• Deleting front before inserting
• Not linking new node to old front

3. Consider the following code snippet for a dequeue using linked list:

Dequeue dq;
dq.insertRear(10);
dq.insertFront(20);
dq.insertRear(30);
dq.deleteFront();
dq.insertFront(40);
dq.deleteRear();
dq.display();

What will be the output of dq.display()?

medium

A. 40 -> 10 -> null

B. 40 -> 20 -> 10 -> null

C. 20 -> 40 -> null

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

Solution

1. Step 1: Trace insertions and deletions step-by-step

   insertRear(10): 10 insertFront(20): 20 -> 10 insertRear(30): 20 -> 10 -> 30 deleteFront(): removes 20, now 10 -> 30 insertFront(40): 40 -> 10 -> 30 deleteRear(): removes 30, now 40 -> 10

2. Step 2: Final list after all operations

   40 -> 10 -> null

3. Final Answer:

   40 -> 10 -> null -> Option A

4. Quick Check:

   Trace insert/delete carefully for final list [OK]

Hint: Trace each operation stepwise to find final list [OK]

Common Mistakes:

• Forgetting to update front or rear after deletion
• Mixing order of insertFront and insertRear
• Assuming deleted elements remain

4. The following code snippet for deleting from rear in a dequeue using linked list has a bug:

void deleteRear() {
  if (rear == nullptr) return;
  Node* temp = front;
  while (temp->next != rear) {
    temp = temp->next;
  }
  delete rear;
  rear = temp;
  rear->next = nullptr;
}

What is the bug in this code?

medium

A. It sets rear->next to non-null value.

B. It deletes the front node instead of rear.

C. It does not handle the case when there is only one node in the dequeue.

D. It causes memory leak by not deleting any node.

Solution

1. Step 1: Analyze the loop for single node case

   If only one node exists, front == rear, so while loop condition temp->next != rear fails immediately, temp = front.

2. Step 2: Check what happens after loop

   Deleting rear node, then setting rear = temp (which is front). But if only one node, temp == rear, so deleting rear deletes the only node, then rear points to deleted memory, causing dangling pointer.

3. Final Answer:

   It does not handle the case when there is only one node in the dequeue. -> Option C

4. Quick Check:

   Single node case needs special handling [OK]

Hint: Check single-node case separately before loop [OK]

Common Mistakes:

• Ignoring single-node edge case
• Assuming loop always runs at least once
• Not updating rear pointer correctly

5. You want to implement a function reverseDequeue() that reverses the order of elements in a dequeue implemented using a linked list. Which approach is best to reverse the dequeue efficiently?

hard

A. Delete all nodes and re-insert them in reverse order.

B. Create a new linked list and insert elements from the original dequeue at the rear one by one.

C. Use an auxiliary stack to push all elements, then pop and insert back into the dequeue.

D. Traverse the linked list once, reversing the next pointers of each node, then swap front and rear pointers.

Solution

1. Step 1: Reverse linked list pointers in one traversal

   Reversing next pointers of each node in one pass efficiently reverses the list.

2. Step 2: Swap front and rear pointers

   After reversal, front and rear pointers must be swapped to reflect new ends.

3. Final Answer:

   Traverse the linked list once, reversing the next pointers of each node, then swap front and rear pointers. -> Option D

4. Quick Check:

   In-place reversal with pointer swap is most efficient [OK]

Hint: Reverse pointers in one pass, then swap front and rear [OK]

Common Mistakes:

• Using extra memory unnecessarily
• Re-inserting nodes instead of reversing pointers
• Not swapping front and rear after reversal