Dequeue Using Linked List in DSA Javascript - Build from Scratch

Dequeue Using Linked List

📖 Scenario: Imagine you are managing a line of people waiting for a ride. People can join the line at the front or the back, and they can also leave from the front or the back. This is like a dequeue (double-ended queue) where you can add or remove items from both ends.

🎯 Goal: You will build a Dequeue using a linked list in JavaScript. You will create the linked list nodes, add configuration for the dequeue size, implement adding and removing from both ends, and finally print the current state of the dequeue.

📋 What You'll Learn

Create a Node class with value and next properties

Create a Dequeue class with front and rear pointers

Add a maxSize property to limit the dequeue size

Implement addFront and addRear methods to add nodes

Implement removeFront and removeRear methods to remove nodes

Print the dequeue elements from front to rear

💡 Why This Matters

🌍 Real World

Dequeue structures are used in real life for managing lines where people can join or leave from both ends, like in ticket counters or call centers.

💼 Career

Understanding linked list based dequeue helps in jobs involving system design, memory management, and implementing efficient data structures.

1

Create Node and Dequeue Classes

Create a Node class with value and next properties. Then create a Dequeue class with front and rear properties initialized to null.

DSA Javascript

# Your code here
// Create Node and Dequeue classes

Hint

Start by defining two classes. The Node holds a value and a pointer to the next node. The Dequeue keeps track of the front and rear nodes.

2

Add maxSize Property to Dequeue

Inside the Dequeue class constructor, add a property called maxSize and set it to 5. Also add a property size initialized to 0 to track current number of nodes.

DSA Javascript

class Node {
    constructor(value) {
        this.value = value;
        this.next = null;
    }
}

class Dequeue {
    constructor() {
        this.front = null;
        this.rear = null;
        // Your code here
    }
}

Hint

Adding maxSize helps limit how many nodes the dequeue can hold. size keeps track of how many nodes are currently inside.

3

Implement addRear and removeFront Methods

Inside the Dequeue class, implement the addRear(value) method to add a new node at the rear if size is less than maxSize. Also implement removeFront() method to remove the node from the front and return its value. Update size accordingly.

DSA Javascript

class Node {
    constructor(value) {
        this.value = value;
        this.next = null;
    }
}

class Dequeue {
    constructor() {
        this.front = null;
        this.rear = null;
        this.maxSize = 5;
        this.size = 0;
    }

    // Your code here
}

Hint

To add at rear, create a new node and link it after the current rear. To remove from front, move the front pointer to the next node and return the old front's value.

4

Print Dequeue Elements from Front to Rear

Create a method printDequeue() inside the Dequeue class that prints all node values from front to rear separated by -> and ending with null. Then create a Dequeue instance, add these values at rear: 10, 20, 30, remove one from front, add 40 at rear, and finally call printDequeue().

DSA Javascript

class Node {
    constructor(value) {
        this.value = value;
        this.next = null;
    }
}

class Dequeue {
    constructor() {
        this.front = null;
        this.rear = null;
        this.maxSize = 5;
        this.size = 0;
    }

    addRear(value) {
        if (this.size >= this.maxSize) {
            return;
        }
        const newNode = new Node(value);
        if (!this.rear) {
            this.front = newNode;
            this.rear = newNode;
        } else {
            this.rear.next = newNode;
            this.rear = newNode;
        }
        this.size++;
    }

    removeFront() {
        if (!this.front) {
            return null;
        }
        const removedValue = this.front.value;
        this.front = this.front.next;
        if (!this.front) {
            this.rear = null;
        }
        this.size--;
        return removedValue;
    }

    // Your code here
}

// Your code here

Hint

Traverse from front to rear, collecting values. Use console.log to print the string. Then test by adding and removing nodes as described.

Practice

1. What is the main advantage of using a linked list to implement a dequeue compared to an array?

easy

A. It automatically sorts the elements as they are inserted.

B. It allows efficient insertion and deletion from both ends without shifting elements.

C. It uses less memory than an array for the same number of elements.

D. It prevents duplicate elements from being added.

Solution

1. Step 1: Understand linked list operations

   A linked list allows adding or removing nodes at the front or back in constant time without moving other elements.

2. Step 2: Compare with array behavior

   Arrays require shifting elements when inserting or deleting at the front, which is less efficient.

3. Final Answer:

   It allows efficient insertion and deletion from both ends without shifting elements. -> Option B

4. Quick Check:

   Linked list = efficient ends operations [OK]

Hint: Linked list avoids shifting elements for front/back operations [OK]

Common Mistakes:

• Thinking arrays are always more memory efficient
• Assuming linked list sorts elements automatically
• Believing linked list prevents duplicates

2. Which of the following is the correct way to add a new node at the front of a dequeue implemented using a linked list in JavaScript?

easy

A. Create new node, set its prev to current tail, update tail to new node.

B. Create new node, set its prev to current head, update head to new node.

C. Create new node, set its next to null, update tail to new node.

D. Create new node, set its next to current head, update head to new node.

Solution

1. Step 1: Understand adding at front in linked list

   To add at front, new node's next points to current head, then head updates to new node.

2. Step 2: Check options for correctness

   Create new node, set its next to current head, update head to new node. matches this logic; others incorrectly use prev or tail.

3. Final Answer:

   Create new node, set its next to current head, update head to new node. -> Option D

4. Quick Check:

   Add front = newNode.next = head; head = newNode [OK]

Hint: New front node points next to old head [OK]

Common Mistakes:

• Confusing prev and next pointers
• Updating tail instead of head when adding front
• Setting next to null incorrectly

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

class Node {
  constructor(value) {
    this.value = value;
    this.next = null;
    this.prev = null;
  }
}

class Dequeue {
  constructor() {
    this.head = null;
    this.tail = null;
  }

  addFront(value) {
    const newNode = new Node(value);
    if (!this.head) {
      this.head = this.tail = newNode;
    } else {
      newNode.next = this.head;
      this.head.prev = newNode;
      this.head = newNode;
    }
  }

  addRear(value) {
    const newNode = new Node(value);
    if (!this.tail) {
      this.head = this.tail = newNode;
    } else {
      this.tail.next = newNode;
      newNode.prev = this.tail;
      this.tail = newNode;
    }
  }

  removeFront() {
    if (!this.head) return null;
    const val = this.head.value;
    this.head = this.head.next;
    if (this.head) this.head.prev = null;
    else this.tail = null;
    return val;
  }
}

const dq = new Dequeue();
dq.addRear(10);
dq.addFront(20);
dq.addRear(30);
const removed = dq.removeFront();
console.log(removed);
console.log(dq.head.value + ' -> ' + dq.tail.value);

What will be the output?

medium

A. 10
20 -> 30

B. 20
30 -> 10

C. 20
10 -> 30

D. 10
30 -> 20

Solution

1. Step 1: Trace insertions

   addRear(10): list is 10 addFront(20): list is 20 -> 10 addRear(30): list is 20 -> 10 -> 30

2. Step 2: Trace removal

   removeFront() removes 20, returns 20. List becomes 10 -> 30.

3. Final Answer:

   20 10 -> 30 -> Option C

4. Quick Check:

   Remove front returns 20; list head=10, tail=30 [OK]

Hint: Track head/tail changes step-by-step [OK]

Common Mistakes:

• Confusing addFront and addRear order
• Forgetting to update prev pointers
• Mixing removed value with current head

4. The following code snippet is intended to remove the rear node from a dequeue implemented using a doubly linked list. What is the error?

removeRear() {
  if (!this.tail) return null;
  const val = this.tail.value;
  this.tail = this.tail.prev;
  this.tail.next = null;
  return val;
}

medium

A. It does not handle the case when the list becomes empty after removal.

B. It incorrectly updates the head pointer instead of tail.

C. It should set this.tail.prev to null instead of this.tail.next.

D. It returns the wrong value after removal.

Solution

1. Step 1: Analyze removal logic

   After setting tail to tail.prev, if tail becomes null (list empty), accessing tail.next causes error.

2. Step 2: Identify missing check

   Code must check if tail is null before setting tail.next = null to avoid runtime error.

3. Final Answer:

   It does not handle the case when the list becomes empty after removal. -> Option A

4. Quick Check:

   Check empty list after removal to avoid null pointer [OK]

Hint: Check for empty list before pointer updates [OK]

Common Mistakes:

• Not checking if tail is null after update
• Confusing head and tail pointers
• Setting wrong pointer to null

5. You want to implement a function reverseDequeue() that reverses the order of nodes in a dequeue implemented using a doubly linked list. Which approach correctly reverses the dequeue?

hard

A. Traverse each node, swap its next and prev pointers, then swap head and tail pointers.

B. Traverse each node, set next to null, and prev to head, then update head to tail.

C. Create a new dequeue and addRear nodes from the original dequeue's front to rear.

D. Swap values of head and tail nodes only.

Solution

1. Step 1: Understand reversal in doubly linked list

   Reversing means each node's next and prev pointers must be swapped to reverse direction.

2. Step 2: Update head and tail pointers

   After swapping pointers in all nodes, head and tail pointers must be swapped to reflect new ends.

3. Final Answer:

   Traverse each node, swap its next and prev pointers, then swap head and tail pointers. -> Option A

4. Quick Check:

   Swap next/prev in nodes + swap head/tail [OK]

Hint: Swap next and prev pointers in all nodes [OK]

Common Mistakes:

• Only swapping head and tail values
• Setting next to null incorrectly
• Creating new dequeue instead of reversing in place