Insert at End of Circular Linked List in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after inserting 4 at the end?

Given a circular linked list with nodes 1 -> 2 -> 3 -> back to 1, what is the printed state after inserting 4 at the end?

DSA Rust

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

struct CircularLinkedList {
    tail: Option<Box<Node>>,
}

impl CircularLinkedList {
    fn new() -> Self {
        CircularLinkedList { tail: None }
    }

    fn insert_end(&mut self, data: i32) {
        let mut new_node = Box::new(Node { data, next: None });
        if self.tail.is_none() {
            new_node.next = Some(new_node.clone());
            self.tail = Some(new_node);
        } else {
            let tail = self.tail.as_mut().unwrap();
            new_node.next = tail.next.take();
            tail.next = Some(new_node);
            self.tail = tail.next.as_mut().map(|node| node.clone());
        }
    }

    fn print_list(&self) {
        if self.tail.is_none() {
            println!("List is empty");
            return;
        }
        let mut temp = self.tail.as_ref().unwrap().next.as_ref();
        loop {
            print!("{} -> ", temp.unwrap().data);
            if temp.unwrap().data == self.tail.as_ref().unwrap().data {
                break;
            }
            temp = temp.unwrap().next.as_ref();
        }
        println!("back to {}", self.tail.as_ref().unwrap().data);
    }
}

fn main() {
    let mut cll = CircularLinkedList::new();
    cll.insert_end(1);
    cll.insert_end(2);
    cll.insert_end(3);
    cll.insert_end(4);
    cll.print_list();
}

A 1 -> 2 -> 3 -> 4 -> back to 1

B 4 -> 1 -> 2 -> 3 -> back to 4

C 1 -> 2 -> 3 -> 4 -> back to 4

D 1 -> 2 -> 3 -> back to 1

💡 Hint

Remember that in a circular linked list, the last node points back to the first node.

After inserting 4 at the end, the list becomes 1 -> 2 -> 3 -> 4, and since it's circular, 4 points back to 1.

🧠 Conceptual

intermediate

Why update the tail pointer when inserting at the end?

In a circular linked list, why do we update the tail pointer after inserting a new node at the end?

A To point to the first node in the list

B To keep track of the last node for efficient insertion at the end

C To delete the first node after insertion

D To reverse the list automatically

💡 Hint

Think about how to quickly add nodes at the end without traversing the whole list.

Updating the tail pointer ensures we always know the last node, making insertion at the end efficient.

🔧 Debug

advanced

Identify the error in this insert_end implementation

What error will this Rust code cause when inserting at the end of a circular linked list?

DSA Rust

fn insert_end(&mut self, data: i32) {
    let mut new_node = Box::new(Node { data, next: None });
    if self.tail.is_none() {
        new_node.next = Some(new_node.clone());
        self.tail = Some(new_node);
    } else {
        let tail = self.tail.as_mut().unwrap();
        new_node.next = tail.next.take();
        tail.next = Some(new_node);
        // Missing update of tail pointer here
    }
}

A Memory leak due to not freeing nodes

B Compilation error due to missing semicolon

C Runtime panic due to unwrap on None

D The list will lose its circular property after insertion

💡 Hint

Check what happens to the tail pointer after insertion.

Without updating the tail pointer, the last node is not updated, breaking the circular link.

📝 Syntax

advanced

Which option correctly sets the new node's next pointer?

In Rust, how do you correctly set the new node's next pointer to the current head in a circular linked list?

A new_node.next = Some(tail.next.unwrap());

B new_node.next = tail.next;

C new_node.next = tail.next.take();

D new_node.next = tail.next.clone();

💡 Hint

Consider ownership and borrowing rules in Rust.

Using take() moves the ownership of the next node to new_node.next safely.

🚀 Application

expert

After multiple insertions, how many nodes are in the circular linked list?

Starting from an empty circular linked list, you insert nodes with values 5, 10, 15, 20 at the end. How many nodes does the list contain?

A 4

B 3

C 5

D 1

💡 Hint

Each insertion adds one node at the end.

Each insert_end call adds one node, so after 4 insertions, there are 4 nodes.

Practice

1. What is the main difference between a circular linked list and a regular singly linked list?

easy

A. In a circular linked list, the last node points back to the first node.

B. In a circular linked list, nodes have two pointers instead of one.

C. In a circular linked list, nodes are stored in an array.

D. In a circular linked list, nodes cannot be inserted at the end.

Solution

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

   A circular linked list connects the last node back to the first node, forming a circle.

2. Step 2: Compare with a regular singly linked list

   A regular singly linked list ends with a node pointing to null, not back to the first node.

3. Final Answer:

   In a circular linked list, the last node points back to the first node. -> Option A

4. Quick Check:

   Circular linked list last node points to head [OK]

Hint: Remember: circular means last node loops to first [OK]

Common Mistakes:

• Thinking circular lists have two pointers per node
• Confusing circular linked list with doubly linked list
• Assuming circular lists cannot have end insertions

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

easy

A. let new_node = Node::new(data); new_node.next = None; tail.next = Some(new_node); tail = new_node;

B. let new_node = Node::new(data); new_node.next = head; tail.next = Some(new_node); tail = new_node;

C. let new_node = Node::new(data); new_node.next = tail; head.next = Some(new_node); tail = new_node;

D. let new_node = Node::new(data); new_node.next = head; head.next = Some(new_node); tail = new_node;

Solution

1. Step 1: Create new node and link to head

   In a circular linked list, the new node's next should point to head to maintain the circle.

2. Step 2: Update tail's next and tail pointer

   Tail's next should point to the new node, and tail should be updated to new node.

3. Final Answer:

   let new_node = Node::new(data); new_node.next = head; tail.next = Some(new_node); tail = new_node; -> Option B

4. Quick Check:

   New node points to head, tail updated [OK]

Hint: New node's next must point to head in circular list [OK]

Common Mistakes:

• Setting new node's next to None breaks the circle
• Linking new node to tail instead of head
• Not updating tail pointer after insertion

3. Given the following Rust code for inserting at the end of a circular linked list, what will be the printed output after inserting 4 into the list 1 -> 2 -> 3 -> (back to 1)?

fn print_list(head: &Option) {
    let mut temp = head.as_ref();
    if temp.is_none() { return; }
    let start = temp.unwrap();
    loop {
        print!("{} -> ", temp.unwrap().data);
        temp = temp.unwrap().next.as_ref();
        if temp.is_none() || temp.unwrap().data == head.as_ref().unwrap().data { break; }
    }
    println!("null");
}

// After insertion of 4 at end
// List structure: 1 -> 2 -> 3 -> 4 -> (back to 1)

medium

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

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

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

D. 4 -> 1 -> 2 -> 3 -> null

Solution

1. Step 1: Understand the print loop logic

   The loop prints nodes until it reaches the head again, then breaks to avoid infinite loop.

2. Step 2: Trace the list after insertion

   After inserting 4 at end, list is 1 -> 2 -> 3 -> 4 -> back to 1. The print stops before repeating head.

3. Final Answer:

   1 -> 2 -> 3 -> 4 -> null -> Option C

4. Quick Check:

   Print stops before repeating head [OK]

Hint: Print stops before repeating head node [OK]

Common Mistakes:

• Printing the head node twice causing infinite loop
• Not breaking loop on head node detection
• Confusing order of nodes after insertion

4. The following Rust code attempts to insert a new node at the end of a circular linked list but causes a runtime error. What is the error?

fn insert_end(tail: &mut Option, data: i32) {
    let mut new_node = Box::new(Node::new(data));
    new_node.next = None;
    if tail.is_none() {
        *tail = Some(new_node);
        tail.as_mut().unwrap().next = tail.clone();
    } else {
        new_node.next = tail.as_ref().unwrap().next.clone();
        tail.as_mut().unwrap().next = Some(new_node);
        *tail = tail.as_mut().unwrap().next.clone();
    }
}

medium

A. Using tail.clone() causes ownership issues leading to runtime panic.

B. Setting new_node.next to None breaks the circular link causing infinite loop.

C. Assigning *tail = tail.as_mut().unwrap().next.clone() causes borrow checker error.

D. Not updating tail pointer after insertion causes stale tail reference.

Solution

1. Step 1: Analyze tail.clone() usage

   Cloning tail (which is an Option) duplicates ownership, which Rust forbids at runtime for Box.

2. Step 2: Understand Rust ownership rules

   Box pointers cannot be cloned freely; this causes runtime panic or compile error due to ownership violation.

3. Final Answer:

   Using tail.clone() causes ownership issues leading to runtime panic. -> Option A

4. Quick Check:

   Cloning Box without proper ownership causes error [OK]

Hint: Avoid cloning Box pointers; use references or Rc instead [OK]

Common Mistakes:

• Setting next to None breaking circularity
• Ignoring Rust ownership and cloning rules
• Not updating tail pointer correctly

5. You have a circular linked list with nodes containing values [10, 20, 30]. You want to insert a new node with value 25 at the end, but also maintain a sorted order in the circular list. Which approach correctly inserts 25 while keeping the list sorted?

hard

A. Replace the last node's value with 25 without changing links.

B. Insert 25 at the end and then sort the entire list.

C. Insert 25 at the head and update tail to new node.

D. Traverse the list to find the correct position and insert 25 there, updating links accordingly.

Solution

1. Step 1: Understand sorted circular linked list insertion

   To keep the list sorted, find the correct position where 25 fits between nodes.

2. Step 2: Insert node at correct position and update links

   Traverse nodes until finding node with value less than 25 and next node greater than 25, then insert new node there.

3. Step 3: Update tail if inserted at end

   If inserted after the current tail, update tail pointer to new node to maintain list integrity.

4. Final Answer:

   Traverse the list to find the correct position and insert 25 there, updating links accordingly. -> Option D

5. Quick Check:

   Insert at sorted position to keep order [OK]

Hint: Insert at correct sorted position, not just at end [OK]

Common Mistakes:

• Inserting at end without sorting breaks order
• Replacing node value loses data integrity
• Not updating tail when inserting at end