Dequeue Using Linked List in DSA Typescript - Time & Space Complexity

Time Complexity: Dequeue Using Linked List

O(1)

Understanding Time Complexity

We want to understand how fast operations on a dequeue using a linked list run as the number of items grows.

Specifically, how the time to add or remove items changes when the dequeue gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

class Dequeue {
  head: Node | null = null;
  tail: Node | null = null;

  addFront(value: number) {
    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;
    }
  }
}
    

This code adds an item to the front of a dequeue implemented with a linked list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Adding a node at the front involves changing a few pointers.
• How many times: Each addFront call does a fixed number of steps, no loops.

How Execution Grows With Input

Adding one item always takes the same small number of steps, no matter how many items are already in the dequeue.

Input Size (n)	Approx. Operations
10	5-6 steps
100	5-6 steps
1000	5-6 steps

Pattern observation: The time stays about the same no matter how big the dequeue grows.

Final Time Complexity

Time Complexity: O(1)

This means adding or removing from the front happens in constant time, quickly and independently of size.

Common Mistake

[X] Wrong: "Adding to the front takes longer as the dequeue grows because we have more items."

[OK] Correct: The linked list keeps track of the front, so adding just changes a few pointers, no matter how many items are inside.

Interview Connect

Knowing that linked list dequeue operations run in constant time shows you understand efficient data structures, a key skill in coding interviews.

Self-Check

"What if we changed the linked list to a simple array? How would the time complexity of adding to the front change?"

Practice

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

easy

A. It allows insertion and deletion from both ends in constant time.

B. It uses less memory than arrays.

C. It automatically sorts the elements.

D. It prevents duplicate elements.

Solution

1. Step 1: Understand dequeue operations

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

2. Step 2: Compare linked list and array

   Linked lists allow adding/removing nodes at both ends in constant time, unlike arrays which may require shifting elements.

3. Final Answer:

   It allows insertion and deletion from both ends in constant time. -> Option A

4. Quick Check:

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

Hint: Linked list supports fast add/remove at both ends [OK]

Common Mistakes:

• Thinking arrays allow constant time insert/delete at front
• Confusing memory usage advantage
• Assuming automatic sorting

2. Which of the following is the correct TypeScript method signature to insert an element at the front of a dequeue implemented using a linked list?

easy

A. insertFront(value: number): void

B. addRear(value: number): number

C. pushFront(): void

D. enqueueFront(value: string): string

Solution

1. Step 1: Identify method purpose and parameter

   Inserting at front requires a method that takes a value (number) and returns nothing (void).

2. Step 2: Match signature with correct naming and types

   insertFront(value: number): void matches: method name 'insertFront', parameter 'value: number', return type 'void'. Others have wrong names, return types, or parameter types.

3. Final Answer:

   insertFront(value: number): void -> Option A

4. Quick Check:

   Insert front method = insertFront(value: number): void [OK]

Hint: Method to add front takes value and returns void [OK]

Common Mistakes:

• Using wrong method names like pushFront
• Wrong return types like number or string
• Incorrect parameter types

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

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

What will be the printed output?

medium

A. 10 -> 20 -> 30 -> null

B. 40 -> 10 -> null

C. 20 -> 10 -> 30 -> null

D. 40 -> 20 -> null

Solution

1. Step 1: Trace each operation on dequeue

   Start empty. 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: Display final dequeue state

   Prints elements from front to rear: 40 -> 10 -> null

3. Final Answer:

   40 -> 10 -> null -> Option B

4. Quick Check:

   After operations, dequeue = 40 -> 10 -> null [OK]

Hint: Trace each insert/delete step carefully [OK]

Common Mistakes:

• Forgetting deleteFront removes front element
• Mixing order of insertFront and insertRear
• Not updating links after deleteRear

4. The following TypeScript method for deleting the rear node in a dequeue implemented using a singly linked list has a bug. What is the bug?

deleteRear(): number | null {
  if (!this.rear) return null;
  let temp = this.front;
  while (temp.next !== this.rear) {
    temp = temp.next;
  }
  const value = this.rear.value;
  this.rear = temp;
  this.rear.next = null;
  return value;
}

medium

A. It causes an infinite loop when the list is empty.

B. It incorrectly updates the front pointer.

C. It returns the wrong value after deletion.

D. It does not handle the case when there is only one node.

Solution

1. Step 1: Analyze loop and pointers

   The loop finds the node before rear by checking temp.next !== rear. If only one node exists, temp.next is null, so loop condition fails immediately.

2. Step 2: Check single node case

   If only one node, front and rear point to same node. The code does not update front or rear to null, causing incorrect state after deletion.

3. Final Answer:

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

4. Quick Check:

   Single node deleteRear needs special handling [OK]

Hint: Check single-node edge case in deleteRear [OK]

Common Mistakes:

• Ignoring single-node list case
• Assuming front pointer updates automatically
• Overlooking null pointer exceptions

5. You want to implement a dequeue using a doubly linked list in TypeScript. Which of the following changes will improve the efficiency of deleting the rear node compared to a singly linked list implementation?

hard

A. Traverse from front to find the node before rear each time.

B. Use an array instead of linked list for rear deletion.

C. Maintain a rear pointer and use the previous pointer to delete rear in O(1) time.

D. Delete rear by shifting all elements forward.

Solution

1. Step 1: Understand doubly linked list structure

   Each node has pointers to next and previous nodes, allowing backward traversal.

2. Step 2: Use rear pointer and previous link

   With rear pointer and previous link, deleting rear node is direct and takes constant time without traversal.

3. Final Answer:

   Maintain a rear pointer and use the previous pointer to delete rear in O(1) time. -> Option C

4. Quick Check:

   Doubly linked list rear delete = O(1) using prev pointer [OK]

Hint: Use prev pointer in doubly linked list for fast rear delete [OK]

Common Mistakes:

• Traversing from front wastes time
• Using array shifts is inefficient
• Ignoring doubly linked list advantages