Delete a Node from Circular Linked List in DSA Rust - Build from Scratch

Delete a Node from Circular Linked List

📋 What You'll Learn

💡 Why This Matters

🌍 Real World

Circular linked lists are used in real-time multiplayer games to manage players in a round-robin fashion, ensuring smooth turns and easy removal of players.

💼 Career

Understanding circular linked lists and node deletion is important for software engineers working on systems requiring cyclic data structures, such as operating systems, network scheduling, and game development.

1

Create the Circular Linked List

Create a struct called Node with fields data: i32 and next: Option<Rc<RefCell<Node>>>. Then create four nodes with values 10, 20, 30, and 40 linked in a circular way. Store the head node in a variable called head. Use Rc and RefCell for shared ownership and mutability.

DSA Rust

// Add necessary use statements
// Define Node struct and create circular linked list with 4 nodes
// Your code here

Hint

Import Rc and RefCell. Define the Node struct. Create Rc<RefCell<Node>> for each node and link them circularly using borrow_mut() and clone().

2

Create the Key Variable

Create a variable called key and set it to 30. This represents the value of the node to delete from the circular linked list.

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;

struct Node {
    data: i32,
    next: Option>>,
}

fn main() {
    let node1 = Rc::new(RefCell::new(Node { data: 10, next: None }));
    let node2 = Rc::new(RefCell::new(Node { data: 20, next: None }));
    let node3 = Rc::new(RefCell::new(Node { data: 30, next: None }));
    let node4 = Rc::new(RefCell::new(Node { data: 40, next: None }));

    node1.borrow_mut().next = Some(node2.clone());
    node2.borrow_mut().next = Some(node3.clone());
    node3.borrow_mut().next = Some(node4.clone());
    node4.borrow_mut().next = Some(node1.clone());

    let head = Some(node1.clone());

    // Create key variable
    // Your code here
}

Hint

Just create a variable key and assign it the value 30.

3

Write the Delete Function

Write a function called delete_node that takes head: Option<Rc<RefCell<Node>>> and key: i32 as parameters and returns Option<Rc<RefCell<Node>>>. This function should delete the node with value key from the circular linked list and return the new head.

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;

struct Node {
    data: i32,
    next: Option>>,
}

fn delete_node(head: Option>>, key: i32) -> Option>> {
    // Your code here
}

fn main() {
    let node1 = Rc::new(RefCell::new(Node { data: 10, next: None }));
    let node2 = Rc::new(RefCell::new(Node { data: 20, next: None }));
    let node3 = Rc::new(RefCell::new(Node { data: 30, next: None }));
    let node4 = Rc::new(RefCell::new(Node { data: 40, next: None }));

    node1.borrow_mut().next = Some(node2.clone());
    node2.borrow_mut().next = Some(node3.clone());
    node3.borrow_mut().next = Some(node4.clone());
    node4.borrow_mut().next = Some(node1.clone());

    let head = Some(node1.clone());

    let key = 30;

    // Call delete_node function
    // Your code here
}

Hint

Use Rc::ptr_eq to detect cycles. Traverse with prev_rc and curr_rc, update prev_rc.next to skip the node with key using borrow_mut(). Handle head deletion by returning the next node as new head.

4

Print the Circular Linked List

Write code to print the circular linked list starting from head in the format: 10 -> 20 -> 40 -> (back to head). Use a loop to traverse until you reach the head again by comparing data.

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;

struct Node {
    data: i32,
    next: Option>>,
}

fn delete_node(head: Option>>, key: i32) -> Option>> {
    if head.is_none() {
        return head;
    }

    let head_rc = head.as_ref().unwrap().clone();
    let head_data = head_rc.borrow().data;

    if head_rc.borrow().next.as_ref().map_or(false, |next_rc| Rc::ptr_eq(next_rc, &head_rc)) && head_data == key {
        return None;
    }

    let mut prev_rc = head_rc.clone();
    let mut curr_rc = head_rc.borrow().next.as_ref().unwrap().clone();

    loop {
        if curr_rc.borrow().data == key {
            let next_next_rc = curr_rc.borrow().next.as_ref().unwrap().clone();
            prev_rc.borrow_mut().next = Some(next_next_rc.clone());
            if Rc::ptr_eq(&curr_rc, &head_rc) {
                return Some(next_next_rc);
            }
            return Some(head_rc.clone());
        }
        prev_rc = curr_rc.clone();
        curr_rc = prev_rc.borrow().next.as_ref().unwrap().clone();
        if Rc::ptr_eq(&curr_rc, &head_rc) {
            return Some(head_rc.clone());
        }
    }
}

fn main() {
    let node1 = Rc::new(RefCell::new(Node { data: 10, next: None }));
    let node2 = Rc::new(RefCell::new(Node { data: 20, next: None }));
    let node3 = Rc::new(RefCell::new(Node { data: 30, next: None }));
    let node4 = Rc::new(RefCell::new(Node { data: 40, next: None }));

    node1.borrow_mut().next = Some(node2.clone());
    node2.borrow_mut().next = Some(node3.clone());
    node3.borrow_mut().next = Some(node4.clone());
    node4.borrow_mut().next = Some(node1.clone());

    let head = Some(node1.clone());

    let key = 30;

    let head = delete_node(head, key);

    if let Some(_) = &head {
        // Your code here
    }
}

Hint

Clone the head Rc. Traverse using current_rc = current_rc.borrow().next.as_ref().unwrap().clone(). Print data with -> until back to start_data, then print (back to head).

Practice

1. What happens when you delete the only node in a circular linked list?

easy

A. The list points to itself indefinitely.

B. The list still contains the deleted node.

C. The list becomes empty (null).

D. The list throws an error.

Solution

1. Step 1: Understand the structure of a single-node circular linked list

   In a circular linked list with one node, the node's next pointer points to itself.

2. Step 2: Delete the only node

   Removing this node means no nodes remain, so the list becomes empty (null).

3. Final Answer:

   The list becomes empty (null). -> Option C

4. Quick Check:

   Single node deleted = empty list [OK]

Hint: Deleting the only node empties the list [OK]

Common Mistakes:

• Assuming the list still contains the deleted node
• Thinking the list points to itself after deletion
• Expecting an error instead of empty list

2. Which Rust syntax correctly declares a mutable circular linked list node struct with a next pointer?

easy

A. struct Node { data: i32, next: Option }

B. struct Node { data: i32, next: &Node }

C. struct Node { data: i32, next: Node }

D. struct Node { data: i32, next: Option> }

Solution

1. Step 1: Understand Rust ownership for circular references

   Rust disallows direct cyclic ownership. Using Rc allows multiple owners and interior mutability.

2. Step 2: Identify correct syntax for mutable circular linked list node

   struct Node { data: i32, next: Option> } uses Option> which supports cycles and mutability safely.

3. Final Answer:

   struct Node { data: i32, next: Option> } -> Option D

4. Quick Check:

   Use Rc for circular mutable nodes [OK]

Hint: Use Rc and RefCell for mutable circular references in Rust [OK]

Common Mistakes:

• Using plain Box<Node> causes ownership errors
• Using references (&Node) disallows ownership
• Using Node directly causes infinite size error

3. Given this Rust code snippet deleting a node with value 3 from a circular linked list:

let mut head = create_circular_list(vec![1, 2, 3, 4]); delete_node(&mut head, 3); print_list(&head);

What is the printed output?

medium

A. 1 -> 2 -> 4 -> 1 (circular)

B. 1 -> 2 -> 3 -> 4 -> 1 (circular)

C. 2 -> 3 -> 4 -> 2 (circular)

D. 1 -> 2 -> 4 -> null

Solution

1. Step 1: Understand the initial list

   The list is circular with nodes 1, 2, 3, 4 linked in a loop.

2. Step 2: Delete node with value 3

   Node 3 is removed, so the list becomes 1 -> 2 -> 4, then back to 1.

3. Final Answer:

   1 -> 2 -> 4 -> 1 (circular) -> Option A

4. Quick Check:

   Deleted 3, list loops 1-2-4 [OK]

Hint: Deleting node removes it, links neighbors circularly [OK]

Common Mistakes:

• Forgetting to link node 2 to node 4
• Printing null instead of circular loop
• Not deleting the correct node

4. Identify the bug in this Rust function snippet for deleting a node from a circular linked list:

fn delete_node(head: &mut Option>, key: i32) { let mut curr = head.clone().unwrap(); while curr.borrow().data != key { curr = curr.borrow().next.clone().unwrap(); } let next_node = curr.borrow().next.clone(); curr.borrow_mut().next = next_node.unwrap().borrow().next.clone(); }

medium

A. It causes infinite loop if key not found.

B. It does not handle deleting the head node correctly.

C. It leaks memory by not freeing nodes.

D. It uses unwrap() unsafely causing panic.

Solution

1. Step 1: Analyze the while loop condition

   The loop continues until curr's data equals key, but no check if key exists, so it loops infinitely if key missing.

2. Step 2: Check for missing exit condition

   Without a break or stop when full circle is done, infinite loop occurs if key not found.

3. Final Answer:

   It causes infinite loop if key not found. -> Option A

4. Quick Check:

   No key check = infinite loop risk [OK]

Hint: Always check for key existence to avoid infinite loops [OK]

Common Mistakes:

• Assuming key always exists
• Not checking for full circle traversal
• Ignoring unwrap() panic risks

5. How would you delete the head node from a circular linked list in Rust while maintaining the circular structure?

hard

A. Replace head data with next node data and delete next node.

B. Find the last node, set its next to head's next, then update head to head's next.

C. Delete head and leave the last node pointing to old head.

D. Set head to None directly without updating other nodes.

Solution

1. Step 1: Locate the last node in the circular list

   The last node points back to head; find it by traversing until next points to head.

2. Step 2: Update last node's next pointer and head

   Set last node's next to head's next node, then update head to head's next to maintain circularity.

3. Final Answer:

   Find last node, link it to head's next, update head to head's next. -> Option B

4. Quick Check:

   Update last node and head to delete head safely [OK]

Hint: Link last node to new head, then update head [OK]

Common Mistakes:

• Setting head to None breaks circularity
• Not updating last node's next pointer
• Deleting head without adjusting links