Stack Implementation Using Linked List in DSA Rust - Time & Space Complexity

Time Complexity: Stack Implementation Using Linked List

O(1)

Understanding Time Complexity

We want to understand how fast the main operations of a stack built with a linked list run.

Specifically, how the time to add or remove items changes as the stack grows.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

struct Stack {
    top: Option>,
}

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

    fn pop(&mut self) -> Option {
        self.top.take().map(|node| {
            self.top = node.next;
            node.data
        })
    }
}
    

This code adds (push) and removes (pop) items at the top of a linked list to behave like a stack.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Adding or removing one node at the top of the linked list.
• How many times: Each push or pop does this once, no loops or traversals involved.

How Execution Grows With Input

Each push or pop works only on the top node, so the time does not increase as the stack grows.

Input Size (n)	Approx. Operations
10	1 operation per push/pop
100	1 operation per push/pop
1000	1 operation per push/pop

Pattern observation: Time stays the same no matter how many items are in the stack.

Final Time Complexity

Time Complexity: O(1)

This means adding or removing an item takes the same short time regardless of stack size.

Common Mistake

[X] Wrong: "Pop or push must take longer as the stack grows because there are more items."

[OK] Correct: We only work with the top item, so the rest of the stack does not slow us down.

Interview Connect

Knowing that stack operations are fast and constant time helps you explain why stacks are good for quick add/remove tasks.

Self-Check

"What if we changed the stack to add items at the end of the linked list instead of the front? How would the time complexity change?"

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