Stack Implementation Using Linked List in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a stack and what principle does it follow?

A stack is a data structure that stores items in a Last In, First Out (LIFO) order. This means the last item added is the first one to be removed.

intermediate

Why use a linked list to implement a stack?

A linked list allows dynamic memory use, so the stack can grow or shrink without a fixed size. It also makes push and pop operations efficient by adding or removing nodes at the head.

intermediate

In Rust, what is the role of the Option type in a linked list stack?

Option is used to represent the possibility of a node being present or absent (None). It helps safely handle the end of the list (null) without unsafe code.

beginner

What does the push operation do in a stack implemented with a linked list?

Push adds a new node at the start (head) of the linked list, making it the new top of the stack.

beginner

How does the pop operation work in a linked list stack?

Pop removes the node at the head of the linked list and returns its value, updating the head to the next node.

What order does a stack follow?

A First In, First Out (FIFO)

B Last In, First Out (LIFO)

C Random Order

D Sorted Order

Stacks follow Last In, First Out (LIFO) order, meaning the last element added is the first to be removed.

In a linked list stack, where is the new element added during push?

A In the middle of the list

B At the tail (end) of the list

C At the head (start) of the list

D At a random position

Push adds the new element at the head of the linked list to keep push and pop operations efficient.

What does the Option type represent in Rust linked list nodes?

A Presence or absence of a node

B The value of the node

C The size of the stack

D The type of data stored

Option is used to represent whether a node exists (Some) or not (None), replacing null pointers safely.

What happens when you pop from an empty stack implemented with a linked list?

A Does nothing silently

B Returns the last element

C Adds a new element

D Returns None or an error indicating empty stack

Popping from an empty stack returns None or an error because there is no element to remove.

Why is a linked list preferred over an array for stack implementation in some cases?

A Because it allows dynamic size without resizing

B Because arrays are slower

C Because linked lists use less memory always

D Because arrays cannot store data

Linked lists allow the stack to grow or shrink dynamically without needing to resize, unlike arrays.

Explain how push and pop operations work in a stack implemented using a linked list.

Think about adding and removing from the start of the list.

Describe the advantages of using a linked list to implement a stack compared to using an array.

Consider memory and operation efficiency.

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