Insert at End of Doubly Linked List in DSA Rust - Time & Space Complexity

Time Complexity: Insert at End of Doubly Linked List

O(n)

Understanding Time Complexity

We want to understand how the time to insert a new item at the end of a doubly linked list changes as the list grows.

How does the number of steps needed grow when the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

fn insert_at_end(head: &mut Option>, value: i32) {
    let mut new_node = Box::new(Node { data: value, prev: None, next: None });
    if head.is_none() {
        *head = Some(new_node);
        return;
    }
    let mut current = head.as_mut().unwrap();
    while current.next.is_some() {
        current = current.next.as_mut().unwrap();
    }
    new_node.prev = Some(&mut **current as *mut Node);
    current.next = Some(new_node);
}
    

This code adds a new node with the given value at the end of a doubly linked list by traversing from the head to the last node.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves through each node from the head to the end.
• How many times: It runs once for each node until the last one, so roughly as many times as the list length.

How Execution Grows With Input

As the list gets longer, the number of steps to reach the end grows directly with the number of nodes.

Input Size (n)	Approx. Operations
10	About 10 steps to reach the end
100	About 100 steps to reach the end
1000	About 1000 steps to reach the end

Pattern observation: The steps increase linearly as the list size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to insert at the end grows in direct proportion to the list size.

Common Mistake

[X] Wrong: "Inserting at the end is always fast and takes constant time."

[OK] Correct: Without a pointer to the last node, you must walk through the whole list, so time grows with list length.

Interview Connect

Understanding how list traversal affects insertion time helps you explain trade-offs in data structure design clearly and confidently.

Self-Check

"What if we kept a pointer to the last node? How would the time complexity change?"

Practice

1. What is the main purpose of inserting a node at the end of a doubly linked list?

easy

A. To reverse the list

B. To remove the first node

C. To add a new element after the last node

D. To find the middle node

Solution

1. Step 1: Understand the structure of a doubly linked list

   A doubly linked list has nodes connected forward and backward, allowing traversal in both directions.

2. Step 2: Purpose of insertion at the end

   Inserting at the end means adding a new node after the current last node, updating pointers accordingly.

3. Final Answer:

   To add a new element after the last node -> Option C

4. Quick Check:

   Insert at end = add after last node [OK]

Hint: Insert at end means add after last node [OK]

Common Mistakes:

• Confusing insertion at end with deletion
• Thinking insertion reverses the list
• Assuming insertion happens at the start

2. Which Rust code snippet correctly creates a new node and inserts it at the end of a doubly linked list?

easy

A. let new_node = Node::new(value); new_node.prev = Some(head); head.next = Some(new_node);

B. let new_node = Node::new(value); head.prev = Some(new_node); new_node.next = Some(head);

C. let new_node = Node::new(value); new_node.next = Some(tail); tail.prev = Some(new_node);

D. let new_node = Node::new(value); tail.next = Some(new_node); new_node.prev = Some(tail);

Solution

1. Step 1: Create new node with given value

   Use Node::new(value) to create a new node.

2. Step 2: Link new node after tail

   Set tail.next to new_node and new_node.prev to tail to insert at end.

3. Final Answer:

   let new_node = Node::new(value); tail.next = Some(new_node); new_node.prev = Some(tail); -> Option D

4. Quick Check:

   New node after tail = let new_node = Node::new(value); tail.next = Some(new_node); new_node.prev = Some(tail); [OK]

Hint: New node's prev points to tail; tail's next points to new node [OK]

Common Mistakes:

• Linking new node before head instead of after tail
• Swapping next and prev pointers incorrectly
• Not updating tail's next pointer

3. Given the Rust code below, what will be the printed list after inserting 4 at the end?

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

// Assume list initially: 1 <-> 2 <-> 3
// Insert 4 at end
// Print list from head to tail

medium

A. 1 <-> 2 <-> 3 <-> 4 <-> null

B. 4 <-> 1 <-> 2 <-> 3 <-> null

C. 1 <-> 2 <-> 4 <-> 3 <-> null

D. 1 <-> 2 <-> 3 <-> null

Solution

1. Step 1: Understand initial list structure

   The list is 1 linked to 2, linked to 3, with 3 as tail.

2. Step 2: Insert 4 at the end

   Adding 4 after 3 updates 3.next to 4 and 4.prev to 3.

3. Final Answer:

   1 <-> 2 <-> 3 <-> 4 <-> null -> Option A

4. Quick Check:

   Insert at end adds 4 after 3 [OK]

Hint: New node appears after last node in printed list [OK]

Common Mistakes:

• Inserting new node at start instead of end
• Not updating next pointer of last node
• Printing list without new node

4. Identify the error in this Rust code snippet for inserting at the end of a doubly linked list:

fn insert_end(tail: &mut Option, value: i32) {
    let mut new_node = Box::new(Node::new(value));
    new_node.prev = tail.clone();
    tail = Some(new_node);
}

medium

A. tail is not updated correctly; should update tail's next pointer

B. new_node.prev should be None for end insertion

C. new_node should not be boxed

D. value should be mutable

Solution

1. Step 1: Check pointer updates

   tail is reassigned locally, but tail's next pointer is not updated to link new_node.

2. Step 2: Correct linking needed

   Must set tail's next to new_node and new_node.prev to tail, then update tail reference.

3. Final Answer:

   tail is not updated correctly; should update tail's next pointer -> Option A

4. Quick Check:

   Update tail.next to new_node [OK]

Hint: Always update tail's next pointer when inserting at end [OK]

Common Mistakes:

• Only reassigning tail without linking nodes
• Setting prev pointer incorrectly
• Ignoring ownership rules in Rust

5. You have a doubly linked list with nodes: 10 <-> 20 <-> 30. You want to insert 25 at the end but also keep a pointer to the middle node for quick access. After insertion, which node should the middle pointer reference to keep it as the middle of the list?

hard

A. 25

B. 20

C. 30

D. 10

Solution

1. Step 1: Understand list before insertion

   List is 10 <-> 20 <-> 30; middle is 20 (second of three nodes).

2. Step 2: Insert 25 at end and find new middle

   List becomes 10 <-> 20 <-> 30 <-> 25. With 4 nodes, middle is between 20 and 30, usually the second node (20) to keep balance.

3. Final Answer:

   20 -> Option B

4. Quick Check:

   Middle of 4 nodes = 2nd node [OK]

Hint: Middle of even nodes is lower middle [OK]

Common Mistakes:

• Moving middle pointer to last inserted node
• Choosing last node as middle
• Not updating middle pointer after insertion