Delete from End of Doubly Linked List in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a doubly linked list?

A doubly linked list is a chain of nodes where each node has two links: one to the next node and one to the previous node. This allows moving forward and backward through the list.

beginner

What happens when you delete the last node from a doubly linked list?

The last node is removed, the previous node's next link is set to null, and the list's tail pointer is updated to the previous node.

intermediate

In Rust, why do we use Option and Box for linked list nodes?

Option is used to represent the possibility of no node (null), and Box is used to allocate nodes on the heap so they can own their data and be linked safely.

beginner

What is the time complexity of deleting the last node in a doubly linked list if you have a tail pointer?

It is O(1) because you can directly access the last node and update pointers without traversing the list.

intermediate

What should you check before deleting the last node in a doubly linked list?

You should check if the list is empty (no nodes) or has only one node, as these cases need special handling.

What pointer is updated when deleting the last node in a doubly linked list?

A The previous node's next pointer

B The head pointer

C The next node's previous pointer

D The first node's previous pointer

When deleting the last node, the previous node's next pointer is set to null to mark the new end of the list.

If a doubly linked list has only one node, what happens after deleting it?

A The list still has one node

B The list becomes empty

C The head points to the deleted node

D The tail points to the deleted node

Deleting the only node makes the list empty, so head and tail pointers become null.

What Rust type is commonly used to represent a nullable pointer in linked lists?

A Box

B Vec

C Result

D Option

Option is used to represent a value that can be Some(node) or None (null).

What is the time complexity of deleting the last node if you do NOT have a tail pointer?

A O(n)

B O(1)

C O(log n)

D O(n^2)

Without a tail pointer, you must traverse the list to find the last node, which takes O(n) time.

Which pointer in the last node of a doubly linked list is always null?

A tail

B prev

C next

D head

The last node's next pointer is null because there is no node after it.

Explain step-by-step how to delete the last node from a doubly linked list.

Think about special cases and pointer updates.

Describe how Rust's ownership and Option types help safely manage nodes in a doubly linked list.

Focus on Rust's memory safety features.

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: Understand the tail node removal

   When deleting the last node, the node before it becomes the new tail.

2. Step 2: Update pointers

   The new tail's next pointer must be set to null to mark the end of the list.

3. Final Answer:

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

4. Quick Check:

   Tail update = new tail with next = null [OK]

Hint: Last node removal means update tail and set next to null [OK]

Common Mistakes:

• Not updating the tail pointer
• Forgetting to set next pointer of new tail to null
• Deleting the head instead of the tail

2. Which Rust code snippet correctly deletes the last node from a doubly linked list?

easy

A. if let Some(tail) = self.tail.take() { if let Some(prev) = tail.prev.take() { prev.next = null; self.tail = Some(prev); } else { self.head = null; } }

B. self.head = null; self.tail = null;

C. self.tail = self.tail.unwrap().prev;

D. self.tail = null; self.head = null;

Solution

1. Step 1: Check tail existence and update pointers

   The code safely takes the tail node, then updates the previous node's next pointer to null.

2. Step 2: Update tail or head accordingly

   If previous node exists, tail is updated to it; else, list becomes empty by setting head to null.

3. Final Answer:

   Code snippet that safely updates tail and head pointers -> Option A

4. Quick Check:

   Safe pointer update with if let Some(tail) = self.tail.take() { if let Some(prev) = tail.prev.take() { prev.next = null; self.tail = Some(prev); } else { self.head = null; } } [OK]

Hint: Use take() and update prev.next and tail safely [OK]

Common Mistakes:

• Directly assigning tail without checking Option
• Not updating previous node's next pointer
• Setting head to null without checking list length

3. Given the following Rust code for a doubly linked list deletion from end, what will be the printed list after deletion?

let mut list = DoublyLinkedList::new();
list.push_back(1);
list.push_back(2);
list.push_back(3);
list.delete_from_end();
list.print();

medium

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

B. 1 -> 2 -> null

C. 2 -> 3 -> null

D. null

Solution

1. Step 1: Initial list state

   The list contains nodes 1, 2, 3 in order.

2. Step 2: Delete last node

   Deleting from end removes node 3, so list becomes 1 -> 2.

3. Final Answer:

   1 -> 2 -> null -> Option B

4. Quick Check:

   After deletion, last node removed = 1 -> 2 [OK]

Hint: Deleting end removes last node only [OK]

Common Mistakes:

• Assuming head node is deleted
• Not updating tail pointer
• Printing old list without deletion

4. Identify the bug in this Rust function that deletes the last node from a doubly linked list:

fn delete_from_end(&mut self) {
    if self.tail.is_none() {
        return;
    }
    let tail = self.tail.unwrap();
    if let Some(prev) = tail.prev {
        prev.next = null;
        self.tail = Some(prev);
    } else {
        self.head = null;
        self.tail = null;
    }
}

medium

A. prev.next is not set to null

B. Not updating head pointer when list has multiple nodes

C. Using unwrap() on Option without safe handling causes panic

D. Function does not check if list is empty

Solution

1. Step 1: Analyze unwrap() usage

   Calling unwrap() on tail without checking can panic if tail is null.

2. Step 2: Suggest safe handling

   Use if let or take() to safely handle Option values without panic.

3. Final Answer:

   Using unwrap() on Option without safe handling causes panic -> Option C

4. Quick Check:

   Unsafe unwrap() = panic risk [OK]

Hint: Avoid unwrap() on Option; use safe pattern matching [OK]

Common Mistakes:

• Ignoring Option safety and causing runtime panic
• Not updating tail pointer after deletion
• Forgetting to set prev.next to null

5. You have a doubly linked list with nodes containing values [10, 20, 30]. After deleting the last node twice, what will be the state of the list?

hard

A. 10 -> 20 -> null

B. null

C. 20 -> 30 -> null

D. 10 -> null

Solution

1. Step 1: Initial list nodes

   The list starts as 10 -> 20 -> 30.

2. Step 2: Delete last node once

   Removing last node 30 results in 10 -> 20.

3. Step 3: Delete last node again

   Removing last node 20 results in 10.

4. Final Answer:

   10 -> null -> Option D

5. Quick Check:

   Two deletions remove last two nodes [OK]

Hint: Each delete removes last node, update tail and next [OK]

Common Mistakes:

• Deleting head instead of tail
• Not updating tail after each deletion
• Assuming list becomes empty too early