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DSA Rustprogramming~5 mins

Delete from Beginning of Doubly Linked List in DSA Rust - Time & Space Complexity

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Time Complexity: Delete from Beginning of Doubly Linked List
O(1)
Understanding Time Complexity

When we delete the first node from a doubly linked list, we want to know how the time it takes changes as the list grows.

We ask: How much work does this deletion need compared to the list size?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


struct Node {
    data: i32,
    prev: Option<*mut Node>,
    next: Option>,
}

fn delete_from_beginning(head: &mut Option>) {
    if let Some(mut node) = head.take() {
        *head = node.next.take();
        if let Some(ref mut new_head) = head {
            new_head.prev = None;
        }
    }
}

This code removes the first node of a doubly linked list and updates the head pointer.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Access and update the head node and its next pointer.
  • How many times: This happens only once per deletion, no loops or recursion.
How Execution Grows With Input

Deleting the first node always takes the same steps, no matter how big the list is.

Input Size (n)Approx. Operations
105-6 steps
1005-6 steps
10005-6 steps

Pattern observation: The number of steps stays the same as the list grows.

Final Time Complexity

Time Complexity: O(1)

This means deleting the first node takes a fixed amount of time, no matter how large the list is.

Common Mistake

[X] Wrong: "Deleting the first node takes longer if the list is bigger because we have to check all nodes."

[OK] Correct: We only update the head and the next node's pointers, no need to visit other nodes.

Interview Connect

Understanding this operation's time helps you explain efficient list manipulations clearly and confidently in interviews.

Self-Check

"What if we changed to deleting from the end of the doubly linked list? How would the time complexity change?"

Practice

(1/5)
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
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