Why Delete from End of Doubly Linked List in DSA Typescript?

The Big Idea

What if you could remove the last item instantly without searching the whole list?

The Scenario

Imagine you have a long chain of paper clips linked together. You want to remove the last paper clip, but you only have access to the first one. You try to find the last clip by counting one by one from the start.

The Problem

Counting each paper clip from the start every time you want to remove the last one is slow and tiring. You might lose track or make mistakes, especially if the chain is very long.

The Solution

A doubly linked list keeps track of both the start and the end of the chain. This way, you can quickly find and remove the last paper clip without counting from the beginning.

Before vs After

✗ Before

let current = head;
while (current.next !== null) {
  current = current.next;
}
// remove current

✓ After

if (tail !== null) {
  let nodeToRemove = tail;
  tail = tail.prev;
  if (tail !== null) {
    tail.next = null;
  } else {
    head = null;
  }
}

What It Enables

This lets you quickly remove the last item in a list, making your programs faster and easier to manage.

Real Life Example

Think of a music playlist where you want to remove the last song you added. Using a doubly linked list, you can instantly find and delete that song without checking every song from the start.

Key Takeaways

Manual removal from the end is slow because you must find the last node first.

Doubly linked lists keep track of the end, allowing quick removal.

This improves speed and reduces errors in managing lists.

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