Why Insert at Beginning of Doubly Linked List in DSA Javascript?

The Big Idea

What if you could add new things to the start of your list instantly without rewriting everything?

The Scenario

Imagine you have a list of your favorite songs written on paper. You want to add a new favorite song right at the top of the list every time you discover one. Doing this by rewriting the whole list every time is tiring and slow.

The Problem

Manually rewriting the entire list to add a new song at the beginning takes a lot of time and can cause mistakes like skipping songs or mixing the order. It's hard to keep track and update the list quickly.

The Solution

Using a doubly linked list lets you add a new item at the start easily by just changing a few links. You don't have to rewrite the whole list, and you can move forward or backward through the list smoothly.

Before vs After

✗ Before

let songs = ['Song1', 'Song2', 'Song3'];
songs = ['NewSong', ...songs];

✓ After

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

// Insert new node at beginning by adjusting pointers

What It Enables

This lets you quickly add new items at the start and navigate both forward and backward through your list without hassle.

Real Life Example

Think of a browser history where you want to add the latest visited page at the front and be able to go back and forth easily.

Key Takeaways

Manually adding at the start is slow and error-prone.

Doubly linked lists let you insert at the beginning by changing just a few links.

This makes adding and navigating items fast and easy.

Practice

1. What is the main purpose of inserting a node at the beginning of a doubly linked list?

easy

A. To add a new node before the current head node

B. To add a new node after the tail node

C. To remove the first node from the list

D. To reverse the entire list

Solution

1. Step 1: Understand the head node position

   The head node is the first node in the doubly linked list.

2. Step 2: Insert new node before head

   Inserting at the beginning means placing the new node before the current head, making it the new head.

3. Final Answer:

   To add a new node before the current head node -> Option A

4. Quick Check:

   Insert at beginning = add before head [OK]

Hint: New node becomes new head in doubly linked list [OK]

Common Mistakes:

• Confusing insertion at beginning with insertion at end
• Thinking insertion removes nodes
• Mixing up head and tail nodes

2. Which of the following is the correct way to update the previous pointer of the old head node when inserting a new node at the beginning in JavaScript?

easy

A. oldHead.prev = newNode;

B. newNode.next = oldHead.prev;

C. newNode.prev = oldHead;

D. oldHead.next = newNode;

Solution

1. Step 1: Identify the old head's previous pointer

   The old head's previous pointer should point to the new node after insertion.

2. Step 2: Update old head's prev to new node

   Setting oldHead.prev = newNode links the old head back to the new node.

3. Final Answer:

   oldHead.prev = newNode; -> Option A

4. Quick Check:

   Old head prev points to new node [OK]

Hint: Old head's prev points to new node after insertion [OK]

Common Mistakes:

• Setting oldHead.next instead of prev
• Confusing newNode's pointers
• Assigning newNode.next incorrectly

3. What will be the printed state of the doubly linked list after inserting a node with value 10 at the beginning of the list 5 <-> 15 <-> 20?

medium

A. 5 <-> 15 <-> 20 <-> 10

B. 10 <-> 15 <-> 20 <-> 5

C. 10 <-> 5 <-> 15 <-> 20

D. 5 <-> 10 <-> 15 <-> 20

Solution

1. Step 1: Insert 10 before current head (5)

   The new node with value 10 becomes the new head, pointing next to 5.

2. Step 2: Link the rest of the list after 5

   The list continues as 5 <-> 15 <-> 20 unchanged after insertion.

3. Final Answer:

   10 <-> 5 <-> 15 <-> 20 -> Option C

4. Quick Check:

   New head 10 followed by old list [OK]

Hint: New node appears first, old list follows [OK]

Common Mistakes:

• Appending new node at end instead of beginning
• Swapping nodes incorrectly
• Ignoring pointer updates

4. Given the following code snippet for inserting at the beginning of a doubly linked list, what is the error?

function insertAtBeginning(head, data) {
  const newNode = { data: data, prev: null, next: head };
  head.prev = newNode;
  head = newNode;
  return head;
}

medium

A. It incorrectly sets newNode.next to null

B. It does not handle empty list when head is null

C. It forgets to update newNode.prev to head

D. It returns the old head instead of new head

Solution

1. Step 1: Check for empty list case

   If head is null, head.prev causes error because head is null.

2. Step 2: Missing null check before setting head.prev

   Code should check if head exists before setting head.prev = newNode.

3. Final Answer:

   It does not handle empty list when head is null -> Option B

4. Quick Check:

   Null head requires special handling [OK]

Hint: Always check if head is null before accessing its properties [OK]

Common Mistakes:

• Ignoring empty list case
• Assuming head is never null
• Not returning updated head

5. You have a doubly linked list with nodes: 7 <-> 14 <-> 21. You want to insert a new node with value 3 at the beginning. After insertion, what will be the correct sequence of node values when traversing from head to tail and tail to head?

hard

A. Head to tail: 21 <-> 14 <-> 7 <-> 3; Tail to head: 3 <-> 7 <-> 14 <-> 21

B. Head to tail: 7 <-> 14 <-> 21 <-> 3; Tail to head: 3 <-> 21 <-> 14 <-> 7

C. Head to tail: 3 <-> 7 <-> 14 <-> 21; Tail to head: 3 <-> 7 <-> 14 <-> 21

D. Head to tail: 3 <-> 7 <-> 14 <-> 21; Tail to head: 21 <-> 14 <-> 7 <-> 3

Solution

1. Step 1: Insert 3 before current head 7

   The new node 3 becomes the head, pointing next to 7.

2. Step 2: Traverse forward and backward

   Forward traversal: 3 -> 7 -> 14 -> 21; backward traversal reverses this order.

3. Final Answer:

   Head to tail: 3 <-> 7 <-> 14 <-> 21; Tail to head: 21 <-> 14 <-> 7 <-> 3 -> Option D

4. Quick Check:

   New head first, tail last, pointers reversed backward [OK]

Hint: Head to tail is forward order; tail to head is reverse [OK]

Common Mistakes:

• Mixing forward and backward traversal order
• Appending new node at tail instead of head
• Incorrect pointer updates causing wrong traversal