Circular vs Linear Linked List Key Difference in DSA Rust - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a linked list is circular by comparing the next pointer of the last node.

DSA Rust

if last_node.next == [1] {
    println!("The list is circular");
} else {
    println!("The list is linear");
}

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

Ahead

BSome(None)

CSome(head)

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Using None instead of head to check circularity.

In a circular linked list, the last node's next points back to the head node.

2fill in blank

medium

Complete the code to traverse a linear linked list until the end (None).

DSA Rust

while let Some(node) = current {
    println!("{}", node.value);
    current = node.[1];
}

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Atail

Bnext

Chead

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' or 'head' instead of 'next' to traverse.

To move forward in a linked list, we follow the next pointer of the current node.

3fill in blank

hard

Fix the error in the code that tries to detect circular linked list by comparing nodes.

DSA Rust

fn is_circular(head: Option<Box<Node>>) -> bool {
    let mut slow = &head;
    let mut fast = &head;
    while fast.is_some() && fast.as_ref().unwrap().next.is_some() {
        slow = &slow.as_ref().unwrap().next;
        fast = &fast.as_ref().unwrap().next.as_ref().unwrap().[1];
        if slow == fast {
            return true;
        }
    }
    false
}

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

Atail

Bprev

Chead

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' or 'head' instead of 'next' for fast pointer.

The fast pointer moves two steps ahead by following 'next' twice to detect cycles.

4fill in blank

hard

Fill both blanks to create a function that returns the last node's next pointer in a linked list.

DSA Rust

fn get_last_next(mut node: &Option<Box<Node>>) -> Option<&Option<Box<Node>>> {
    while node.as_ref().unwrap().[1].is_some() {
        node = &node.as_ref().unwrap().[2];
    }
    Some(&node.as_ref().unwrap().next)
}

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

Anext

Bprev

Chead

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' or 'head' instead of 'next' to traverse.

To find the last node, keep moving through 'next' until it is None.

5fill in blank

hard

Fill all three blanks to create a function that builds a circular linked list from a vector of values.

DSA Rust

fn build_circular_list(values: Vec<i32>) -> Option<Box<Node>> {
    let mut head = None;
    let mut tail = None;
    for val in values {
        let new_node = Box::new(Node { value: val, next: None });
        if head.is_none() {
            head = Some(new_node);
            tail = head.as_mut();
        } else {
            tail.unwrap().[1] = Some(new_node);
            tail = tail.unwrap().[2].as_mut();
        }
    }
    if let Some(t) = tail {
        t.[3] = head.clone();
    }
    head
}

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

Anext

Bprev

Cclone

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' instead of 'next' to link nodes.

Not linking last node to head.

We link nodes using 'next', and finally point the last node's 'next' to head to make it circular.

Practice

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

easy

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

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

C. In a linear linked list, nodes point to multiple next nodes.

D. In a linear linked list, the last node points back to the first node.

Solution

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

   A linear linked list has nodes connected one after another, and the last node points to null, meaning the list ends.

2. Step 2: Understand the structure of a circular linked list

   In a circular linked list, the last node points back to the first node, creating a loop with no null end.

3. Final Answer:

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

4. Quick Check:

   Circular list loops back = D [OK]

Hint: Circular loops back to start; linear ends with null [OK]

Common Mistakes:

• Thinking circular lists store nodes in arrays
• Believing linear lists loop back to start
• Confusing multiple next nodes with circular structure

2. Which of the following Rust code snippets correctly creates a circular linked list node pointing to itself?

easy

A. let mut node = Node { data: 5, next: None }; node.next = Some(Box::new(node.clone()));

B. let node = Node { data: 5, next: None };

C. let mut node = Node { data: 5, next: None }; node.next = Some(Box::new(node));

D. let node = Node { data: 5, next: Some(Box::new(node)) };

Solution

1. Step 1: Understand Rust ownership and self-referencing

   Rust does not allow direct self-referencing without cloning or using smart pointers carefully.

2. Step 2: Analyze each option for correctness

   let node = Node { data: 5, next: None }; has next as None, so no circular link. let node = Node { data: 5, next: Some(Box::new(node)) }; tries to use node before it's fully created, causing error. let mut node = Node { data: 5, next: None }; node.next = Some(Box::new(node)); tries to assign node.next to node itself, which is invalid due to ownership. let mut node = Node { data: 5, next: None }; node.next = Some(Box::new(node.clone())); clones node and assigns it to next, creating a circular reference.

3. Final Answer:

   let mut node = Node { data: 5, next: None }; node.next = Some(Box::new(node.clone())); -> Option A

4. Quick Check:

   Clone node for circular link = B [OK]

Hint: Clone node to point next to itself in Rust [OK]

Common Mistakes:

• Assigning next before node is fully created
• Using None for next in circular list
• Ignoring Rust ownership rules

3. Given the following Rust code for a circular linked list with three nodes, what will be printed when traversing starting from head and printing data 4 times?

#[derive(Clone)]
struct Node {
    data: i32,
    next: Option<Box<Node>>,
}

fn main() {
    let node3 = Node { data: 3, next: None };
    let node2 = Node { data: 2, next: Some(Box::new(node3)) };
    let mut node1 = Node { data: 1, next: Some(Box::new(node2)) };
    
    // Making circular
    let cloned_head = node1.clone();
    node1.next.as_mut().unwrap().next.as_mut().unwrap().next = Some(Box::new(cloned_head));

    let mut current = &node1;
    for _ in 0..4 {
        println!("{}", current.data);
        current = current.next.as_ref().unwrap();
    }
}

medium

A. 1 2 3 None

B. 1 2 3 3

C. 1 2 3 2

D. 1 2 3 1

Solution

1. Step 1: Understand the linked list structure

   Nodes 1, 2, 3 are linked linearly first. Then node3.next points back to the first node, making it circular.

2. Step 2: Trace the traversal for 4 prints

   Starting at node1: print 1, move to node2: print 2, move to node3: print 3, move to node1 again (due to circular): print 1.

3. Final Answer:

   1 2 3 1 -> Option D

4. Quick Check:

   Circular list loops back to start = 1 2 3 1 [OK]

Hint: Circular list repeats nodes after last one [OK]

Common Mistakes:

• Assuming last node points to None
• Printing wrong node data on loop
• Confusing node references in Rust

4. Identify the error in this Rust code snippet that tries to create a circular linked list:

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

fn main() {
    let mut node1 = Node { data: 1, next: None };
    let mut node2 = Node { data: 2, next: None };
    node1.next = Some(Box::new(node2));
    node2.next = Some(Box::new(node1));
}

medium

A. Node struct fields are private and inaccessible

B. Missing semicolon after node1.next assignment

C. Cannot assign node2.next after moving node2 into node1.next

D. Rust does not allow Box pointers in structs

Solution

1. Step 1: Understand ownership and moves in Rust

   When node2 is moved into node1.next via Box::new(node2), node2 is no longer accessible in main.

2. Step 2: Identify the error in node2.next assignment

   Trying to assign node2.next after moving node2 causes a compile-time error because node2 is moved and cannot be used again.

3. Final Answer:

   Cannot assign node2.next after moving node2 into node1.next -> Option C

4. Quick Check:

   Ownership moved node2, cannot use again = C [OK]

Hint: Moved value can't be used again in Rust [OK]

Common Mistakes:

• Ignoring Rust ownership and move rules
• Assuming variables remain usable after move
• Confusing Box pointer usage

5. You want to implement a function in Rust that detects if a linked list is circular or linear. Which approach correctly uses the key difference to detect a circular linked list?

hard

A. Count nodes and if count exceeds 10, assume circular.

B. Traverse nodes and check if any node's next pointer points back to the head node.

C. Check if the last node's next pointer is None to confirm circularity.

D. Use recursion to print all nodes until a null pointer is found.

Solution

1. Step 1: Recall the key difference between circular and linear lists

   In circular lists, the last node points back to the head or some earlier node, forming a loop. In linear lists, the last node points to None.

2. Step 2: Choose the correct detection method

   Checking if any node's next points back to the head detects circularity. Counting nodes or checking for None incorrectly assumes circularity or linearity. Recursion may cause infinite loops on circular lists.

3. Final Answer:

   Traverse nodes and check if any node's next pointer points back to the head node. -> Option B

4. Quick Check:

   Loop back to head means circular = A [OK]

Hint: Detect loop by checking if next points to head [OK]

Common Mistakes:

• Assuming node count alone detects circularity
• Checking last node's next for None means circular
• Using recursion without loop detection