Delete from End of Doubly Linked List in DSA Typescript - Build from Scratch

Delete from End of Doubly Linked List

📖 Scenario: You are managing a playlist of songs using a doubly linked list. Each song is a node with a title. You want to remove the last song from the playlist.

🎯 Goal: Build a doubly linked list with three songs, then delete the last song from the list, and finally print the playlist from start to end.

📋 What You'll Learn

Create a doubly linked list with nodes containing song titles: 'Song1', 'Song2', 'Song3'.

Add a variable called head pointing to the first node.

Write a function called deleteFromEnd that removes the last node from the list.

Print the playlist from start to end showing the song titles separated by ' <-> '.

💡 Why This Matters

🌍 Real World

Doubly linked lists are used in music players, browsers, and other apps to move forward and backward through items efficiently.

💼 Career

Understanding linked lists and how to modify them is important for software developers working with data structures and memory management.

1

Create the Doubly Linked List

Create a class called Node with properties title (string), prev (Node or null), and next (Node or null). Then create three nodes with titles 'Song1', 'Song2', and 'Song3'. Link them to form a doubly linked list. Finally, create a variable called head that points to the first node.

DSA Typescript

// Your code here

Hint

Define the Node class with title, prev, and next. Create three nodes and link them by setting next and prev properties. Set head to the first node.

2

Add a Variable for the Current Node

Create a variable called current and set it equal to head. This will help us find the last node.

DSA Typescript

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

const node1 = new Node('Song1');
const node2 = new Node('Song2');
const node3 = new Node('Song3');

node1.next = node2;
node2.prev = node1;
node2.next = node3;
node3.prev = node2;

const head = node1;

// Your code here

Hint

Set current to head so you can move through the list starting from the first node.

3

Write the Function to Delete from End

Write a function called deleteFromEnd that takes no arguments. Inside, use a while loop with current.next !== null to move current to the last node. Then, if current.prev exists, set current.prev.next to null to remove the last node. If current.prev does not exist, set head to null because the list had only one node.

DSA Typescript

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

const node1 = new Node('Song1');
const node2 = new Node('Song2');
const node3 = new Node('Song3');

node1.next = node2;
node2.prev = node1;
node2.next = node3;
node3.prev = node2;

const head = node1;

let current = head;

// Your code here

Hint

Use a while loop to reach the last node. Then remove it by changing the previous node's next to null. If the list has only one node, set head to null.

4

Print the Playlist After Deletion

Call the function deleteFromEnd(). Then create a variable called temp and set it to head. Use a while loop to traverse the list from temp. Collect the title of each node in an array called titles. After the loop, print the titles joined by ' <-> '.

DSA Typescript

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

const node1 = new Node('Song1');
const node2 = new Node('Song2');
const node3 = new Node('Song3');

node1.next = node2;
node2.prev = node1;
node2.next = node3;
node3.prev = node2;

let head = node1;

function deleteFromEnd() {
    let current = head;
    if (current === null) return;
    while (current.next !== null) {
        current = current.next;
    }
    if (current.prev !== null) {
        current.prev.next = null;
    } else {
        head = null;
    }
}

// Your code here

Hint

Call deleteFromEnd() to remove the last song. Then traverse from head to collect titles. Print them joined by ' <-> '.

Practice

1. What happens to the tail pointer when you delete the last node from a doubly linked list?

easy

A. It becomes null immediately

B. It moves to the previous node of the deleted node

C. It points to the head node

D. It remains unchanged

Solution

1. Step 1: Understand the tail pointer role

   The tail pointer always points to the last node in the list.

2. Step 2: Effect of deleting the last node

   When the last node is deleted, the tail must move to the previous node to remain the last node.

3. Final Answer:

   It moves to the previous node of the deleted node -> Option B

4. Quick Check:

   Tail updates to previous node after deletion [OK]

Hint: Tail always points to last node, so move it back after deletion [OK]

Common Mistakes:

• Setting tail to null without checking if list is empty
• Not updating tail pointer after deletion
• Confusing head and tail pointers
• Assuming tail points to head after deletion

2. Which of the following is the correct TypeScript syntax to delete the last node from a doubly linked list assuming tail is not null?

easy

A. tail = tail.prev; tail.next = null;

B. tail = tail.next; tail.prev = null;

C. tail = null; head = null;

D. tail.prev = tail; tail.next = null;

Solution

1. Step 1: Move tail pointer to previous node

   To delete the last node, tail must point to its previous node: tail = tail.prev;

2. Step 2: Set new tail's next to null

   Since tail is now last node, its next must be null: tail.next = null;

3. Final Answer:

   tail = tail.prev; tail.next = null; -> Option A

4. Quick Check:

   Correct syntax moves tail back and nullifies next [OK]

Hint: Tail moves back, then tail.next set to null [OK]

Common Mistakes:

• Moving tail forward instead of backward
• Not setting tail.next to null
• Setting tail to null without updating links
• Confusing prev and next pointers

3. Given the doubly linked list: 1 <-> 2 <-> 3 <-> 4 <-> null, what is the list after deleting the last node?

medium

A. 1 <-> 2 <-> 3 <-> null

B. 1 <-> 2 <-> 3 <-> 4 <-> null

C. 2 <-> 3 <-> 4 <-> null

D. 1 <-> 2 <-> null

Solution

1. Step 1: Identify the last node

   The last node is 4, pointed by tail.

2. Step 2: Delete last node and update tail

   Remove node 4, update tail to node 3, set tail.next to null.

3. Final Answer:

   1 <-> 2 <-> 3 <-> null -> Option A

4. Quick Check:

   Last node 4 removed, tail now 3 [OK]

Hint: Remove last node, tail moves to previous node [OK]

Common Mistakes:

• Not updating tail pointer
• Leaving deleted node linked
• Removing wrong node
• Confusing head and tail updates

4. Consider this TypeScript code snippet to delete the last node from a doubly linked list:

if (tail !== null) {
  tail = tail.prev;
  tail.next = null;
}

What error can occur here?

medium

A. No error, code is correct

B. Syntax error due to missing semicolon

C. Runtime error if list has only one node

D. Tail pointer moves forward incorrectly

Solution

1. Step 1: Analyze single-node list case

   If list has one node, tail.prev is null, so tail becomes null.

2. Step 2: Accessing tail.next causes error

   After tail = tail.prev (null), next line tries tail.next = null, causing runtime error (cannot access property of null).

3. Final Answer:

   Runtime error if list has only one node -> Option C

4. Quick Check:

   Null tail causes runtime error on next access [OK]

Hint: Check if tail.prev is null before setting tail.next [OK]

Common Mistakes:

• Not handling single-node list edge case
• Assuming tail is never null after update
• Ignoring runtime null pointer errors
• Confusing syntax errors with runtime errors

5. You have a doubly linked list with nodes: 10 <-> 20 <-> 30. After deleting the last node twice, what is the final list?

hard

A. 10 <-> 20 <-> null

B. 20 <-> null

C. null

D. 10 <-> null

Solution

1. Step 1: Delete last node first time

   Last node 30 is removed, tail moves to 20, list becomes 10 <-> 20 <-> null.

2. Step 2: Delete last node second time

   Last node 20 is removed, tail moves to 10, list becomes 10 <-> null.

3. Step 3: Verify final list

   After two deletions, list is 10 <-> null.

4. Final Answer:

   10 <-> null -> Option D

5. Quick Check:

   Two deletions remove 30 and 20, leaving 10 [OK]

Hint: Delete last node twice removes last two nodes [OK]

Common Mistakes:

• Deleting only one node instead of two
• Confusing node values after deletion
• Not updating tail pointer correctly
• Assuming list becomes empty too early