Queue Implementation Using Linked List in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node with given value.

DSA Javascript

class Node {
  constructor(value) {
    this.value = [1];
    this.next = null;
  }
}

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

Avalue

Bnext

Cnull

Dthis

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning this.next instead of this.value

Using null instead of the passed value

The constructor assigns the passed value to this.value to store the node's data.

2fill in blank

medium

Complete the enqueue method to add a new node at the end of the queue.

DSA Javascript

enqueue(value) {
  const newNode = new Node(value);
  if (this.rear === null) {
    this.front = newNode;
    this.rear = newNode;
  } else {
    this.rear.[1] = newNode;
    this.rear = newNode;
  }
}

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

Aprev

Bvalue

Cfront

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning value instead of next

Using front or prev which are incorrect here

The rear node's next pointer should point to the new node to link it at the end.

3fill in blank

hard

Fix the error in the dequeue method to correctly remove the front node.

DSA Javascript

dequeue() {
  if (this.front === null) {
    return null;
  }
  const removedValue = this.front.value;
  this.front = this.front.[1];
  if (this.front === null) {
    this.rear = null;
  }
  return removedValue;
}

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

Anext

Bprev

Cvalue

Dfront

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev or value instead of next

Assigning front to front which causes no change

To remove the front node, move the front pointer to front.next.

4fill in blank

hard

Fill both blanks to check if the queue is empty.

DSA Javascript

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

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

Afront

Brear

Cnull

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Checking rear instead of front

Comparing to undefined instead of null

The queue is empty if front is null, meaning no nodes are present.

5fill in blank

hard

Fill all three blanks to implement the peek method that returns the front value without removing it.

DSA Javascript

peek() {
  if (this.[1] === [2]) {
    return null;
  }
  return this.front.[3];
}

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

Afront

Bnull

Cvalue

Drear

Attempts:

3 left

💡 Hint

Common Mistakes

Checking rear instead of front

Returning front instead of front.value

Peek returns the value of the front node if it exists; otherwise, null.

Practice

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

easy

A. Simpler syntax for insertion and deletion

B. Faster random access to elements

C. Uses less memory than arrays

D. Dynamic size without fixed capacity limits

Solution

1. Step 1: Understand queue size limitations in arrays

   Arrays have fixed size or require resizing, which can be costly.

2. Step 2: Linked list dynamic size advantage

   Linked lists grow and shrink dynamically without fixed capacity limits.

3. Final Answer:

   Dynamic size without fixed capacity limits -> Option D

4. Quick Check:

   Linked list queues grow dynamically [OK]

Hint: Linked lists grow dynamically, arrays have fixed size [OK]

Common Mistakes:

• Thinking linked lists allow faster random access
• Assuming linked lists always use less memory
• Confusing syntax simplicity with performance

2. Which of the following is the correct way to add a new node to the end of a linked list queue in JavaScript?

easy

A. tail.next = newNode; tail = newNode;

B. head.next = newNode; head = newNode;

C. newNode.next = tail; tail = newNode;

D. newNode.next = head; head = newNode;

Solution

1. Step 1: Identify tail pointer usage in queue enqueue

   New nodes are added at the tail; tail.next points to new node.

2. Step 2: Update tail to new node

   After adding, tail must point to the new node to keep queue end updated.

3. Final Answer:

   tail.next = newNode; tail = newNode; -> Option A

4. Quick Check:

   Enqueue adds at tail [OK]

Hint: Enqueue adds at tail: tail.next = newNode; tail = newNode [OK]

Common Mistakes:

• Adding new node at head instead of tail
• Not updating tail pointer after insertion
• Confusing head and tail pointers

3. You want to implement a method peek() for a linked list queue that returns the front element without removing it. Which is the correct implementation?

easy

A. return this.head ? this.head.value : null;

B. return this.tail ? this.tail.value : null;

C. return this.head.next ? this.head.next.value : null;

D. return this.tail.next ? this.tail.next.value : null;

Solution

1. Step 1: Understand peek operation in queue

   Peek returns front element without removing it, which is at head in linked list queue.

2. Step 2: Check correct pointer for front element

   Head points to front node; tail points to end node. So return head.value if head exists.

3. Final Answer:

   return this.head ? this.head.value : null; -> Option A

4. Quick Check:

   Peek returns head value [OK]

Hint: Peek returns head value without dequeue [OK]

Common Mistakes:

• Returning tail value instead of head
• Accessing next of tail which is always null
• Returning head.next value which skips front

4. Consider this JavaScript code for a linked list queue:

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

class Queue {
  constructor() {
    this.head = null;
    this.tail = null;
  }
  enqueue(value) {
    const newNode = new Node(value);
    if (!this.tail) {
      this.head = newNode;
      this.tail = newNode;
      return;
    }
    this.tail.next = newNode;
    this.tail = newNode;
  }
  dequeue() {
    if (!this.head) return null;
    const val = this.head.value;
    this.head = this.head.next;
    if (!this.head) this.tail = null;
    return val;
  }
}

const q = new Queue();
q.enqueue(10);
q.enqueue(20);
q.dequeue();
q.enqueue(30);
q.dequeue();
q.enqueue(40);
q.enqueue(50);
q.dequeue();
console.log(q.head.value);

What will be printed?

medium

A. 20

B. 40

C. 50

D. 30

Solution

1. Step 1: Trace enqueue and dequeue operations

   Queue after operations: - enqueue 10: head=10, tail=10 - enqueue 20: head=10->20, tail=20 - dequeue: removes 10, head=20 - enqueue 30: head=20->30, tail=30 - dequeue: removes 20, head=30 - enqueue 40: head=30->40, tail=40 - enqueue 50: head=30->40->50, tail=50 - dequeue: removes 30, head=40

2. Step 2: Identify current head value

   After last dequeue, head points to node with value 40.

3. Final Answer:

   40 -> Option B

4. Quick Check:

   Final head value after operations = 40 [OK]

Hint: Track head after each dequeue to find current front [OK]

Common Mistakes:

• Forgetting to update tail when queue becomes empty
• Mixing up enqueue and dequeue order
• Assuming head value is unchanged after dequeue

5. The following dequeue method for a linked list queue has a bug:

dequeue() {
  if (!this.head) return null;
  const val = this.head.value;
  this.head = this.head.next;
  return val;
}

What is the problem?

medium

A. It removes the tail instead of the head

B. It returns the wrong value

C. It does not update tail when queue becomes empty

D. It causes infinite loop

Solution

1. Step 1: Analyze dequeue method behavior

   It removes head by moving head pointer to next node and returns old head value.

2. Step 2: Identify missing tail update

   If after removal head becomes null (queue empty), tail still points to old node causing inconsistency.

3. Final Answer:

   It does not update tail when queue becomes empty -> Option C

4. Quick Check:

   Empty queue must reset tail = null [OK]

Hint: Reset tail to null when head becomes null after dequeue [OK]

Common Mistakes:

• Ignoring tail pointer update on empty queue
• Confusing head and tail roles
• Assuming dequeue removes tail node