Delete from Beginning of Doubly Linked List in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to update the head pointer after deletion.

DSA Rust

self.head = [1];

Drag options to blanks, or click blank then click option'

Aself.tail

Bself.head.next

CNone

Dself.head.prev

Attempts:

3 left

💡 Hint

Common Mistakes

Setting head to tail instead of next node

Setting head to None without checking list state

After deleting the first node, the head should point to the next node.

2fill in blank

medium

Complete the code to remove the backward link from the new head.

DSA Rust

if let Some(ref mut node) = self.head {
    node.[1] = None;
}

Drag options to blanks, or click blank then click option'

Ahead

Bnext

Ctail

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Setting next pointer instead of prev

Not updating the prev pointer at all

The previous pointer of the new head should be None to detach the old first node.

3fill in blank

hard

Fix the error in the code that handles empty list after deletion.

DSA Rust

if self.head.is_none() {
    self.[1] = None;
}

Drag options to blanks, or click blank then click option'

Aprev

Bhead

Ctail

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Setting head to None again

Not updating tail pointer

If the list becomes empty after deletion, tail should also be None.

4fill in blank

hard

Fill both blanks to correctly detach the first node and update the list.

DSA Rust

if let Some(mut old_head) = self.head.take() {
    self.head = old_head.[1];
    if let Some(ref mut new_head) = self.head {
        new_head.[2] = None;
    }
}

Drag options to blanks, or click blank then click option'

Anext

Bprev

Chead

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing next and prev pointers

Not detaching old head properly

We move head to old_head.next and set new_head.prev to None to detach the old first node.

5fill in blank

hard

Fill all three blanks to complete the delete from beginning function safely.

DSA Rust

pub fn delete_from_beginning(&mut self) {
    if self.head.is_none() {
        return;
    }
    if let Some(mut old_head) = self.head.take() {
        self.head = old_head.[1];
        if let Some(ref mut new_head) = self.head {
            new_head.[2] = None;
        } else {
            self.[3] = None;
        }
    }
}

Drag options to blanks, or click blank then click option'

Anext

Bprev

Ctail

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Not updating tail when list becomes empty

Confusing prev and next pointers

Move head to old_head.next, set new_head.prev to None, and if list is empty, set tail to None.

Practice

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

easy

A. It points to the last node

B. It moves to the next node in the list

C. It becomes null immediately

D. It remains unchanged

Solution

1. Step 1: Understand the role of head in a doubly linked list

   The head pointer always points to the first node in the list.

2. Step 2: Effect of deleting the first node

   When the first node is deleted, the head must move to the next node to keep the list accessible.

3. Final Answer:

   It moves to the next node in the list -> Option B

4. Quick Check:

   Deleting first node updates head to next node [OK]

Hint: Head always points to first node; update it after deletion [OK]

Common Mistakes:

• Not updating head pointer after deletion
• Setting head to null without checking list length
• Confusing head with tail pointer

2. Which of the following Rust code snippets correctly deletes the first node from a doubly linked list?

easy

A. if let Some(mut node) = head.take() { head = node.next.take(); if let Some(ref mut new_head) = head { new_head.prev = None; } }

B. head = head.next; head.prev = None;

C. head = None; tail = None;

D. head.prev = None; head = head.next;

Solution

1. Step 1: Analyze safe Rust deletion pattern

   if let Some(mut node) = head.take() { head = node.next.take(); if let Some(ref mut new_head) = head { new_head.prev = None; } } uses take() to safely remove the first node and updates head and prev pointers correctly.

2. Step 2: Check other options for errors

   head = head.next; head.prev = None; tries to access head.next without handling Option. head = None; tail = None; deletes entire list incorrectly. head.prev = None; head = head.next; accesses head.prev before updating head, causing error.

3. Final Answer:

   Option A code snippet -> Option A

4. Quick Check:

   Safe pointer update with take() = if let Some(mut node) = head.take() { head = node.next.take(); if let Some(ref mut new_head) = head { new_head.prev = None; } } [OK]

Hint: Use take() to safely remove and update pointers [OK]

Common Mistakes:

• Ignoring Option unwrap safety
• Not updating prev pointer of new head
• Setting head to None without checking

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

struct Node { val: i32, prev: Option, next: Option }

// List: 10 <-> 20 <-> 30
let mut head = Some(Box::new(Node { val: 10, prev: None, next: Some(Box::new(Node { val: 20, prev: None, next: Some(Box::new(Node { val: 30, prev: None, next: None })) })) }));

// Delete first node
if let Some(mut node) = head.take() {
  head = node.next.take();
  if let Some(ref mut new_head) = head {
    new_head.prev = None;
  }
}

// Print list
let mut current = &head;
while let Some(node) = current {
  print!("{} -> ", node.val);
  current = &node.next;
}
print!("null");

medium

A. null

B. 10 -> 20 -> 30 -> null

C. 20 -> 30 -> null

D. 30 -> null

Solution

1. Step 1: Initial list structure

   The list starts as 10 <-> 20 <-> 30, with head at 10.

2. Step 2: Deletion of first node

   Deleting first node moves head to 20, and sets 20.prev to None.

3. Step 3: Printing the list

   Starting from new head (20), printing nodes gives 20 -> 30 -> null.

4. Final Answer:

   20 -> 30 -> null -> Option C

5. Quick Check:

   Head updated to second node, list prints from 20 [OK]

Hint: After deletion, head points to second node; print from there [OK]

Common Mistakes:

• Printing old head after deletion
• Not updating prev pointer causing broken links
• Confusing next and prev pointers

4. Identify the error in this Rust code snippet that tries to delete the first node from a doubly linked list:

if head.is_some() {
  head = head.unwrap().next;
  if let Some(ref mut new_head) = head {
    new_head.prev = None;
  }
}

medium

A. Using unwrap() consumes head, causing a move error

B. Not setting new_head.prev to None

C. Missing check if head is None before deletion

D. Incorrectly updating tail pointer

Solution

1. Step 1: Understand unwrap() behavior on Option

   Calling unwrap() on head moves the value out, consuming it and invalidating head.

2. Step 2: Consequence of move error

   Assigning head = head.unwrap().next; causes a move error because head is used after being moved.

3. Final Answer:

   Using unwrap() consumes head, causing a move error -> Option A

4. Quick Check:

   unwrap() moves value, invalidating head [OK]

Hint: Avoid unwrap() on Option when reassigning; use take() [OK]

Common Mistakes:

• Using unwrap() causing move errors
• Not handling Option safely
• Forgetting to update prev pointer

5. You have a doubly linked list with nodes: 5 <-> 15 <-> 25 <-> 35. After deleting the first node twice using correct Rust code, what will be the final list printed?

let mut head = Some(Box::new(Node { val: 5, prev: None, next: Some(Box::new(Node { val: 15, prev: None, next: Some(Box::new(Node { val: 25, prev: None, next: Some(Box::new(Node { val: 35, prev: None, next: None })) })) })) }));

// Delete first node twice
for _ in 0..2 {
  if let Some(mut node) = head.take() {
    head = node.next.take();
    if let Some(ref mut new_head) = head {
      new_head.prev = None;
    }
  }
}

// Print list
let mut current = &head;
while let Some(node) = current {
  print!("{} -> ", node.val);
  current = &node.next;
}
print!("null");

hard

A. 15 -> 25 -> 35 -> null

B. null

C. 35 -> null

D. 25 -> 35 -> null

Solution

1. Step 1: Initial list and first deletion

   List starts as 5 <-> 15 <-> 25 <-> 35. First deletion removes 5, head moves to 15.

2. Step 2: Second deletion

   Second deletion removes 15, head moves to 25, and 25.prev is set to None.

3. Step 3: Printing final list

   Starting from 25, printing nodes gives 25 -> 35 -> null.

4. Final Answer:

   25 -> 35 -> null -> Option D

5. Quick Check:

   Two deletions move head twice forward [OK]

Hint: Each deletion moves head to next node; repeat for multiple deletions [OK]

Common Mistakes:

• Not updating prev pointer after each deletion
• Deleting only once instead of twice
• Printing old head after deletions