Dequeue Using Linked List in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to add a new node at the front of the dequeue.

DSA Javascript

this.front = new Node(data);
this.front.next = [1];

Drag options to blanks, or click blank then click option'

Anull

Bthis.rear

Cthis.front.next

Dthis.front

Attempts:

3 left

💡 Hint

Common Mistakes

Setting next to this.front causes a circular reference.

Setting next to this.rear is incorrect because rear is the end of the list.

When adding a new node at the front of an empty dequeue, its next pointer should point to null to start the list.

2fill in blank

medium

Complete the code to remove a node from the rear of the dequeue.

DSA Javascript

if (this.front === this.rear) {
  this.front = this.rear = null;
} else {
  let current = this.front;
  while (current.next !== [1]) {
    current = current.next;
  }
  current.next = null;
  this.rear = current;
}

Drag options to blanks, or click blank then click option'

Athis.front

Bcurrent

Cnull

Dthis.rear

Attempts:

3 left

💡 Hint

Common Mistakes

Using current.next !== this.front will never find the node before rear.

Using current.next !== null will stop too early.

To remove the rear node, we find the node before rear by checking current.next !== this.rear.

3fill in blank

hard

Fix the error in the code to correctly add a node at the rear of the dequeue.

DSA Javascript

let newNode = new Node(data);
if (this.rear === null) {
  this.front = this.rear = newNode;
} else {
  this.rear.next = [1];
  this.rear = newNode;
}

Drag options to blanks, or click blank then click option'

Athis.front

Bnull

CnewNode

Dthis.rear

Attempts:

3 left

💡 Hint

Common Mistakes

Setting rear.next to null disconnects the list.

Setting rear.next to front causes a cycle.

When adding at the rear, the current rear's next should point to the new node.

4fill in blank

hard

Fill both blanks to check if the dequeue is empty and return the correct boolean.

DSA Javascript

isEmpty() {
  return this.front === [1] && this.rear === [2];
}

Drag options to blanks, or click blank then click option'

Anull

Bundefined

C0

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Checking against undefined is incorrect here.

Checking against 0 or false is not valid for pointers.

The dequeue is empty if both front and rear are null.

5fill in blank

hard

Fill all three blanks to correctly print all elements from front to rear.

DSA Javascript

print() {
  let current = this.front;
  while (current !== [1]) {
    console.log(current.[2]);
    current = current.[3];
  }
}

Drag options to blanks, or click blank then click option'

Anull

Bdata

Cnext

Dfront

Attempts:

3 left

💡 Hint

Common Mistakes

Using front instead of null causes infinite loop.

Printing next instead of data prints wrong value.

We loop until current is null, print current.data, and move to current.next.

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