Queue Implementation Using Linked List in DSA Javascript - Time & Space Complexity

Time Complexity: Queue Implementation Using Linked List

O(1)

Understanding Time Complexity

We want to understand how the time to add or remove items from a queue changes as the queue grows.

How does the queue's size affect the speed of enqueue and dequeue operations?

Scenario Under Consideration

Analyze the time complexity of the following queue operations using a linked list.

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

class Queue {
  constructor() {
    this.front = null;
    this.rear = null;
  }

  enqueue(value) {
    const newNode = new Node(value);
    if (!this.rear) {
      this.front = this.rear = newNode;
      return;
    }
    this.rear.next = newNode;
    this.rear = newNode;
  }

  dequeue() {
    if (!this.front) return null;
    const temp = this.front;
    this.front = this.front.next;
    if (!this.front) this.rear = null;
    return temp.value;
  }
}
    

This code adds items to the end and removes items from the front of the queue using a linked list.

Identify Repeating Operations

Look for loops or repeated steps inside enqueue and dequeue.

• Primary operation: Adding or removing one node by changing pointers.
• How many times: Each operation does a fixed number of steps, no loops.

How Execution Grows With Input

Adding or removing one item always takes the same number of steps, no matter how big the queue is.

Input Size (n)	Approx. Operations
10	5-6 steps per enqueue or dequeue
100	5-6 steps per enqueue or dequeue
1000	5-6 steps per enqueue or dequeue

Pattern observation: The number of steps stays the same regardless of queue size.

Final Time Complexity

Time Complexity: O(1)

This means adding or removing an item takes the same short time no matter how many items are in the queue.

Common Mistake

[X] Wrong: "Removing an item from the queue takes longer as the queue grows because it has to move through all items."

[OK] Correct: The linked list keeps track of the front, so removing just changes a pointer without walking through the list.

Interview Connect

Knowing that queue operations are fast and constant time helps you explain efficient data handling in real projects.

Self-Check

"What if we used an array instead of a linked list for the queue? How would the time complexity of dequeue change?"

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