Stack Implementation Using Linked List in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the Node struct with a value and a next pointer.

DSA Rust

struct Node<T> {
    value: T,
    next: [1],
}

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

ANode<T>

BBox<Node<T>>

COption<Box<Node<T>>>

DOption<Node<T>>

Attempts:

3 left

💡 Hint

Common Mistakes

Using Box> without Option causes infinite size error.

Using Node directly causes recursive type without indirection.

The next pointer in a linked list node is an optional Box pointing to the next Node.

2fill in blank

medium

Complete the push method to add a new node at the top of the stack.

DSA Rust

impl<T> Stack<T> {
    fn push(&mut self, value: T) {
        let new_node = Box::new(Node {
            value,
            next: [1],
        });
        self.top = Some(new_node);
    }
}

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

Aself.top

Bself.top.clone()

CNone

Dself.top.take()

Attempts:

3 left

💡 Hint

Common Mistakes

Using self.top.clone() causes unnecessary cloning.

Using self.top directly causes borrowing errors.

We take ownership of the current top node to link it as next of the new node.

3fill in blank

hard

Fix the error in the pop method to correctly remove and return the top value.

DSA Rust

impl<T> Stack<T> {
    fn pop(&mut self) -> Option<T> {
        self.top.take().map(|node| {
            self.top = [1];
            node.value
        })
    }
}

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

Anode.next

Bnode.next.take()

Cself.top

DSome(node.next)

Attempts:

3 left

💡 Hint

Common Mistakes

Using node.next.take() causes error because node is moved.

Assigning self.top to self.top causes no change.

After taking the top node, set self.top to node.next to update the stack.

4fill in blank

hard

Fill both blanks to implement the peek method returning a reference to the top value.

DSA Rust

impl<T> Stack<T> {
    fn peek(&self) -> Option<&T> {
        self.top.as_ref().map(|node| [1])
    }

    fn peek_mut(&mut self) -> Option<&mut T> {
        self.top.as_mut().map(|node| [2])
    }
}

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

A&node.value

B&mut node.value

Cnode.value

D*node.value

Attempts:

3 left

💡 Hint

Common Mistakes

Returning node.value moves the value out, causing errors.

Using *node.value tries to dereference a value, which is incorrect.

peek returns a reference, so use &node.value; peek_mut returns mutable reference, so use &mut node.value.

5fill in blank

hard

Fill all three blanks to implement the is_empty method and a new Stack constructor.

DSA Rust

impl<T> Stack<T> {
    fn new() -> Self {
        Stack { top: [1] }
    }

    fn is_empty(&self) -> bool {
        self.top [2] None
    }

    fn size(&self) -> usize {
        let mut count = 0;
        let mut current = self.top.as_ref();
        while let Some(node) = current {
            count += 1;
            current = node.[3];
        }
        count
    }
}

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

ANone

B==

Cnext.as_ref()

Dis_some()

Attempts:

3 left

💡 Hint

Common Mistakes

Using is_some() in is_empty returns opposite result.

Using next without as_ref() causes type mismatch.

new initializes top as None; is_empty checks if top == None; size traverses nodes using next.as_ref().

Practice

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

easy

A. It allows random access to elements

B. It uses less memory than an array

C. The stack can grow dynamically without a fixed size limit

D. It is easier to sort the stack elements

Solution

1. Step 1: Understand stack size limitation in arrays

   Arrays have fixed size, so stacks implemented with arrays have a maximum capacity.

2. Step 2: Linked list dynamic size property

   Linked lists can grow or shrink dynamically by adding or removing nodes, so stack size is not fixed.

3. Final Answer:

   The stack can grow dynamically without a fixed size limit -> Option C

4. Quick Check:

   Dynamic size = Advantage of linked list stack [OK]

Hint: Linked list stacks grow dynamically, arrays don't [OK]

Common Mistakes:

• Thinking linked list uses less memory always
• Confusing stack with queue behavior
• Assuming random access is possible in linked list

2. Which of the following Rust code snippets correctly defines a node struct for a linked list stack?

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 and recursive types

   Recursive types like linked lists require indirection using Box and Option for nullable pointers.

2. Step 2: Analyze each option

   struct Node { data: i32, next: Option } uses Option which allows nullable next pointer; struct Node { data: i32, next: Node } and struct Node { data: i32, next: &Node } are invalid due to recursive type without indirection; struct Node { data: i32, next: Box<Node> } misses Option, so no null pointer.

3. Final Answer:

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

4. Quick Check:

   Option = correct linked list pointer [OK]

Hint: Use Option for nullable next in Rust [OK]

Common Mistakes:

• Omitting Option causing infinite size error
• Using references (&Node) without lifetime
• Using Box<Node> without Option for nullable

3. Consider the following Rust code for pushing elements onto a linked list stack:

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

struct Stack {
    head: Option,
}

impl Stack {
    fn push(&mut self, val: i32) {
        let new_node = Box::new(Node { data: val, next: self.head.take() });
        self.head = Some(new_node);
    }
}

fn main() {
    let mut stack = Stack { head: None };
    stack.push(10);
    stack.push(20);
    stack.push(30);
    // What is the top element's data now?
    println!("{}", stack.head.as_ref().unwrap().data);
}

medium

A. 30

B. 20

C. 10

D. None

Solution

1. Step 1: Trace push operations

   Push 10: head points to node with data 10; Push 20: head points to node with data 20, next is 10; Push 30: head points to node with data 30, next is 20.

2. Step 2: Identify top element

   After pushes, top element is the head node's data, which is 30.

3. Final Answer:

   30 -> Option A

4. Quick Check:

   Last pushed = top element = 30 [OK]

Hint: Top is last pushed element in stack [OK]

Common Mistakes:

• Confusing order of nodes in linked list
• Assuming head points to bottom element
• Not using take() causing borrow errors

4. The following Rust code for popping an element from a linked list stack has an error. What is the error?

impl Stack {
    fn pop(&mut self) -> Option<i32> {
        if self.head.is_none() {
            return None;
        }
        let node = self.head.unwrap();
        self.head = node.next;
        Some(node.data)
    }
}

medium

A. Missing lifetime annotations on struct

B. Not checking if stack is empty before pop

C. Returning wrong data type from pop

D. Using unwrap() on Option without handling ownership

Solution

1. Step 1: Analyze unwrap() usage on self.head

   unwrap() consumes the Option, but self.head is borrowed mutably, causing ownership error.

2. Step 2: Correct way is to use take() to move ownership

   Using self.head.take() moves ownership out safely, allowing access to node and updating head.

3. Final Answer:

   Using unwrap() on Option without handling ownership -> Option D

4. Quick Check:

   unwrap() consumes Option, use take() instead [OK]

Hint: Use take() to move Option ownership safely [OK]

Common Mistakes:

• Using unwrap() causing borrow checker errors
• Ignoring empty stack check
• Confusing references and ownership

5. You want to implement a function reverse_stack that reverses the elements of a linked list stack without using extra data structures. Which approach correctly achieves this in Rust?

hard

A. Iteratively pop elements and push them onto a new stack, then replace original stack

B. Recursively pop elements, reverse the rest, then push popped element back

C. Traverse linked list and swap data values between nodes from head to tail

D. Use a vector to collect elements, reverse vector, then rebuild stack

Solution

1. Step 1: Understand constraints - no extra data structures allowed

   Options A and D use extra stacks or vectors, so they violate constraints.

2. Step 2: Recursive approach explanation

   Recursively pop elements until stack is empty, then push elements back in reverse order, effectively reversing stack without extra storage.

3. Step 3: Swapping data values is complex and error-prone

   Traverse linked list and swap data values between nodes from head to tail requires traversal and swapping data, which is complicated and not idiomatic in Rust linked lists.

4. Final Answer:

   Recursively pop elements, reverse the rest, then push popped element back -> Option B

5. Quick Check:

   Recursion reverses stack without extra structures [OK]

Hint: Use recursion to reverse stack without extra storage [OK]

Common Mistakes:

• Using extra data structures despite constraints
• Trying to swap node data directly
• Ignoring ownership rules in recursion