Queue Implementation Using Linked List in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new empty queue.

DSA Rust

fn new() -> Self {
    Queue { front: None, rear: None, size: [1] }
}

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

ANone

B0

C1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting size to 1 instead of 0

Using None instead of 0 for size

The queue starts empty, so size should be 0.

2fill in blank

medium

Complete the code to add a new node at the rear of the queue.

DSA Rust

fn enqueue(&mut self, value: i32) {
    let new_node = Box::new(Node { data: value, next: None });
    if self.rear.is_none() {
        self.front = Some(new_node);
        self.rear = Some(new_node);
    } else {
        self.rear.as_mut().unwrap().next = Some(new_node);
        self.rear = [1];
    }
    self.size += 1;
}

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

Aself.rear.as_mut().unwrap().next.as_mut().unwrap()

Bself.rear.as_ref().unwrap().next.as_mut().unwrap()

Cself.rear.as_ref().unwrap()

Dself.rear.as_mut().unwrap().next

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning rear to the old rear instead of the new node

Using as_ref() instead of as_mut() causing immutable borrow

After linking the new node, rear should point to the new node, which is the next of the old rear. Since rear is an Option>, we need to get a mutable reference to the next node (which is Some(new_node)) and unwrap it to assign rear.

3fill in blank

hard

Fix the error in the dequeue method to correctly remove the front node.

DSA Rust

fn dequeue(&mut self) -> Option<i32> {
    match self.front.take() {
        None => None,
        Some(node) => {
            self.front = [1];
            if self.front.is_none() {
                self.rear = None;
            }
            self.size -= 1;
            Some(node.data)
        }
    }
}

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

Aself.rear

Bnode.next.as_ref()

Cself.front.as_ref()

Dnode.next

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning front to rear incorrectly

Using as_ref() causing type mismatch

After removing the front node, front should point to the next node of the removed node.

4fill in blank

hard

Fill both blanks to implement a method that checks if the queue is empty.

DSA Rust

fn is_empty(&self) -> bool {
    self.size [1] 0 && self.front [2] None
}

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

A==

B!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using != instead of == causing wrong logic

Using || instead of && for condition

The queue is empty if size equals zero and front is None.

5fill in blank

hard

Fill all three blanks to implement a method that returns the front value without removing it.

DSA Rust

fn peek(&self) -> Option<i32> {
    match &self.front {
        Some(node) => Some(node.[1]),
        None => [2],
    }
}

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

Adata

BNone

CSome(0)

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Returning Some(0) instead of None when empty

Accessing next instead of data

Peek returns the data of the front node if it exists, otherwise None.

Practice

1. What is the main advantage of implementing a Queue using a linked list instead of an array?

easy

A. Simpler to implement than arrays

B. Faster access to elements by index

C. Dynamic size without fixed capacity limit

D. Uses less memory for small queues

Solution

1. Step 1: Understand queue size behavior

   Arrays have fixed size or require resizing, linked lists grow dynamically.

2. Step 2: Compare memory and resizing

   Linked lists allocate nodes as needed, so no fixed capacity limit exists.

3. Final Answer:

   Dynamic size without fixed capacity limit -> Option C

4. Quick Check:

   Linked list queues grow dynamically = D [OK]

Hint: Linked lists grow dynamically, arrays have fixed size [OK]

Common Mistakes:

• Thinking linked lists allow fast random access
• Assuming arrays use less memory always
• Confusing implementation simplicity

2. Which of the following is the correct Rust syntax to define a node struct for a linked list queue?

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: Box<Node> }

Solution

1. Step 1: Understand Rust ownership for linked lists

   Rust requires heap allocation for recursive types using Box and Option for nullable links.

2. Step 2: Analyze options

   struct Node { data: i32, next: Option } uses Option which is correct for nullable next pointer; others cause infinite size or borrowing issues.

3. Final Answer:

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

4. Quick Check:

   Use Option for nullable next in Rust [OK]

Hint: Use Option for next pointer in Rust linked list [OK]

Common Mistakes:

• Using direct Node type causes infinite size error
• Using references (&Node) without lifetime causes errors
• Omitting Option for nullable next pointer

3. Given the following Rust code snippet for a queue implemented with a linked list, what will be printed after enqueueing 10, 20 and then dequeueing once?

let mut queue = Queue::new();
queue.enqueue(10);
queue.enqueue(20);
let val = queue.dequeue();
println!("Dequeued: {:?}", val);
queue.print();

medium

A. Dequeued: Some(10)
20 -> null

B. Dequeued: Some(20)
10 -> null

C. Dequeued: None
10 -> 20 -> null

D. Dequeued: Some(10)
10 -> 20 -> null

Solution

1. Step 1: Trace enqueue operations

   Enqueue 10 adds node with 10, then enqueue 20 adds node with 20 at the end.

2. Step 2: Trace dequeue operation

   Dequeue removes front node (10), returns Some(10). Remaining queue has 20 only.

3. Final Answer:

   Dequeued: Some(10) 20 -> null -> Option A

4. Quick Check:

   First in first out, dequeue returns 10, queue left with 20 [OK]

Hint: Dequeue removes front, so first enqueued is first out [OK]

Common Mistakes:

• Assuming dequeue removes last element
• Printing queue before dequeue
• Confusing Some and None return values

4. Identify the bug in this Rust enqueue method for a linked list queue:

fn enqueue(&mut self, data: i32) {
    let new_node = Box::new(Node { data, next: None });
    if self.tail.is_none() {
        self.head = Some(new_node);
    } else {
        self.tail.as_mut().unwrap().next = Some(new_node);
    }
    self.tail = Some(new_node);
}

medium

A. Not updating head when queue is empty

B. Using Box instead of Rc for nodes

C. Missing return statement in enqueue

D. Assigning new_node to tail after moving it causes use-after-move error

Solution

1. Step 1: Analyze ownership of new_node

   new_node is moved into head or tail.next, so cannot be moved again.

2. Step 2: Identify use-after-move error

   Assigning self.tail = Some(new_node) after moving new_node causes compile error.

3. Final Answer:

   Assigning new_node to tail after moving it causes use-after-move error -> Option D

4. Quick Check:

   Moving new_node twice causes error [OK]

Hint: Don't move ownership twice; clone or restructure [OK]

Common Mistakes:

• Ignoring Rust ownership rules
• Not updating tail pointer correctly
• Confusing Box and Rc usage

5. You want to implement a method peek for your linked list queue in Rust that returns the front element without removing it. Which approach is correct?

hard

A. Return Option<i32> by cloning data from head node

B. Return Option<&i32> by borrowing data from head node

C. Return i32 directly, panicking if queue is empty

D. Return Option by moving head node

Solution

1. Step 1: Understand peek method goal

   Peek returns front element without removing it, so no ownership transfer.

2. Step 2: Choose correct return type

   Borrowing data with Option<&i32> avoids cloning or moving, safe and efficient.

3. Final Answer:

   Return Option<&i32> by borrowing data from head node -> Option B

4. Quick Check:

   Peek borrows front data, no ownership change [OK]

Hint: Peek borrows front data, don't move or clone unnecessarily [OK]

Common Mistakes:

• Moving head node causing queue modification
• Returning owned data causing unnecessary clone
• Panicking instead of returning Option