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

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 to remove the last one is slow and tiring. It's easy to 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

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

✓ After

if (tail != null) {
  tail = tail.previous;
  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 program 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 jump straight to the last song and remove it instantly.

Key Takeaways

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

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

This operation improves efficiency and reduces errors.

Practice

1. What happens when you delete the last node from a doubly linked list that has multiple nodes?

easy

A. The second last node becomes the new tail and its next pointer is set to null.

B. The head pointer is set to null.

C. The list becomes circular.

D. The first node is deleted instead.

Solution

1. Step 1: Identify the last node and its previous node

   In a doubly linked list, the last node's previous node is the second last node.

2. Step 2: Update the second last node's next pointer

   Set the second last node's next pointer to null, making it the new tail.

3. Final Answer:

   The second last node becomes the new tail and its next pointer is set to null. -> Option A

4. Quick Check:

   Deleting last node updates tail's next to null [OK]

Hint: Tail's previous node next pointer becomes null after deletion [OK]

Common Mistakes:

• Setting head to null instead of updating tail
• Not updating the previous node's next pointer
• Deleting the first node by mistake

2. Which of the following Java code snippets correctly deletes the last node from a doubly linked list named dll?

easy

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

B. dll.head = dll.head.next; dll.head.prev = null;

C. dll.tail = dll.tail.prev; dll.tail.next = null;

D. dll.head.prev = dll.tail; dll.tail = dll.tail.next;

Solution

1. Step 1: Move tail pointer to previous node

   Set dll.tail to dll.tail.prev to remove last node reference.

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

   Update dll.tail.next to null to mark end of list.

3. Final Answer:

   dll.tail = dll.tail.prev; dll.tail.next = null; -> Option C

4. Quick Check:

   Correct tail update and next pointer null [OK]

Hint: Tail moves back, next pointer set to null [OK]

Common Mistakes:

• Updating head instead of tail
• Setting tail to null without updating pointers
• Confusing next and prev pointers

3. Consider the following Java code for deleting the last node from a doubly linked list:

class Node {
  int data;
  Node prev, next;
  Node(int d) { data = d; prev = next = null; }
}

class DoublyLinkedList {
  Node head, tail;

  void deleteFromEnd() {
    if (tail == null) return;
    if (head == tail) {
      head = tail = null;
      return;
    }
    tail = tail.prev;
    tail.next = null;
  }
}

DoublyLinkedList dll = new DoublyLinkedList();
dll.head = new Node(1);
Node second = new Node(2);
dll.head.next = second;
second.prev = dll.head;
dll.tail = second;
dll.deleteFromEnd();
System.out.println(dll.head.data + " -> " + (dll.head.next == null ? "null" : dll.head.next.data));

What is the output?

medium

A. 1 -> 2

B. null -> null

C. 2 -> null

D. 1 -> null

Solution

1. Step 1: Setup list with two nodes 1 and 2

   Head points to node 1, tail points to node 2, linked properly.

2. Step 2: Delete last node (node 2)

   Tail moves to node 1, tail.next set to null, removing node 2.

3. Final Answer:

   1 -> null -> Option D

4. Quick Check:

   After deletion, only node 1 remains with next null [OK]

Hint: After deleting tail, tail.prev.next is null [OK]

Common Mistakes:

• Printing old tail node data
• Not updating tail.next to null
• Confusing head and tail pointers

4. The following code attempts to delete the last node from a doubly linked list but causes a runtime error:

void deleteFromEnd() {
  if (tail == null) return;
  tail = tail.prev;
  tail.next = null;
}

What is the error and how to fix it?

medium

A. Compilation error due to missing semicolon; fix by adding semicolon.

B. NullPointerException when list has one node; fix by checking if tail.prev is null.

C. Logical error deleting head instead of tail; fix by updating head pointer.

D. No error; code works fine.

Solution

1. Step 1: Identify cause of NullPointerException

   If list has one node, tail.prev is null, so tail becomes null, then tail.next causes error.

2. Step 2: Fix by checking single node case

   Add condition: if head == tail, set head and tail to null before updating pointers.

3. Final Answer:

   NullPointerException when list has one node; fix by checking if tail.prev is null. -> Option B

4. Quick Check:

   Check single node case before pointer update [OK]

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

Common Mistakes:

• Not handling single node list
• Assuming tail.prev always exists
• Ignoring null pointer exceptions

5. You have a doubly linked list with nodes: 10 <-> 20 <-> 30 <-> 40. After deleting the last node twice using the correct deleteFromEnd method, what is the final list?

hard

A. 10 <-> 20

B. 10 <-> 20 <-> 30

C. 20 <-> 30

D. 30 <-> 40

Solution

1. Step 1: Delete last node once

   Deleting last node 40 removes it; list becomes 10 <-> 20 <-> 30.

2. Step 2: Delete last node second time

   Deleting last node 30 removes it; list becomes 10 <-> 20.

3. Final Answer:

   10 <-> 20 -> Option A

4. Quick Check:

   Two deletions remove last two nodes [OK]

Hint: Each deleteFromEnd removes tail node [OK]

Common Mistakes:

• Deleting from head instead of tail
• Not updating tail pointer after deletion
• Confusing node values after deletion