Dynamic Stack Using Resizable Array in DSA Rust - Build from Scratch

Dynamic Stack Using Resizable Array

📖 Scenario: You are building a simple stack data structure that can grow dynamically as you add more items. Imagine a stack of plates that can get taller when you add more plates.

🎯 Goal: Create a dynamic stack using a resizable array (Rust Vec) that supports pushing and popping elements.

📋 What You'll Learn

Create a stack using a Vec<i32> named stack with initial elements

Add a variable capacity to track the current capacity of the stack

Write code to push a new element onto the stack and increase capacity if needed

Write code to pop the top element from the stack

Print the final state of the stack after operations

💡 Why This Matters

🌍 Real World

Dynamic stacks are used in many applications like undo features in text editors, expression evaluation, and backtracking algorithms.

💼 Career

Understanding dynamic data structures like stacks is essential for software development roles involving algorithms, system design, and performance optimization.

1

Create the initial stack

Create a Vec<i32> called stack with these exact elements: 10, 20, 30

DSA Rust

# Create a Vec called stack with elements 10, 20, 30
# Your code here

Hint

Use vec![10, 20, 30] to create the vector and name it stack.

2

Add capacity variable

Add a variable called capacity and set it to the current capacity of stack using stack.capacity()

DSA Rust

fn main() {
    let mut stack: Vec = vec![10, 20, 30];
    // Add capacity variable here
    # Your code here
}

Hint

Use stack.capacity() to get the current capacity and store it in capacity.

3

Push and pop elements

Push the element 40 onto stack. If the length of stack is now greater than capacity, update capacity to the new stack.capacity(). Then pop the top element from stack and store it in a variable called popped.

DSA Rust

fn main() {
    let mut stack: Vec = vec![10, 20, 30];
    let mut capacity = stack.capacity();
    // Push 40 and update capacity if needed
    // Pop the top element and store in popped
    # Your code here
}

Hint

Use stack.push(40) to add, check length with stack.len(), update capacity if needed, then use stack.pop() to remove the top element.

4

Print the final stack

Print the final state of stack using println!("{:?}", stack)

DSA Rust

fn main() {
    let mut stack: Vec = vec![10, 20, 30];
    let mut capacity = stack.capacity();
    stack.push(40);
    if stack.len() > capacity {
        capacity = stack.capacity();
    }
    let popped = stack.pop();
    // Print the stack
    # Your code here
}

Hint

Use println!("{:?}", stack) to print the vector in debug format.

Practice

1. What is the main advantage of using a dynamic stack implemented with a resizable array compared to a fixed-size array stack?

easy

A. It does not require resizing or copying elements.

B. It can grow or shrink in size as needed during runtime.

C. It always has a fixed maximum size.

D. It uses less memory than a linked list stack.

Solution

1. Step 1: Understand fixed vs dynamic size

   A fixed-size array stack has a set capacity that cannot change during runtime, limiting the number of elements it can hold.

2. Step 2: Analyze dynamic stack behavior

   A dynamic stack using a resizable array can increase or decrease its capacity by resizing the underlying array, allowing it to handle more or fewer elements as needed.

3. Final Answer:

   It can grow or shrink in size as needed during runtime. -> Option B

4. Quick Check:

   Dynamic resizing = flexible size [OK]

Hint: Dynamic stack resizes array to fit elements [OK]

Common Mistakes:

• Thinking dynamic stack has fixed size
• Confusing linked list with array stack
• Assuming no resizing needed

2. Which Rust code snippet correctly doubles the capacity of a vector when the stack is full?

easy

A. if stack.len() == stack.capacity() { stack.reserve(stack.capacity()); }

B. if stack.len() == stack.capacity() { stack.reserve(1); }

C. if stack.len() == stack.capacity() { stack.reserve(stack.len() / 2); }

D. if stack.len() == stack.capacity() { stack.reserve(0); }

Solution

1. Step 1: Understand vector capacity and reserve method

   The vector's capacity is the total space allocated. To double capacity, we add capacity() more space using reserve(capacity()).

2. Step 2: Check each option's reserve argument

   if stack.len() == stack.capacity() { stack.reserve(stack.capacity()); } reserves current capacity again, doubling it. Others reserve too little or zero, not doubling.

3. Final Answer:

   if stack.len() == stack.capacity() { stack.reserve(stack.capacity()); } -> Option A

4. Quick Check:

   Reserve current capacity to double [OK]

Hint: Reserve current capacity to double vector size [OK]

Common Mistakes:

• Reserving fixed small number instead of doubling
• Using zero in reserve causing no growth
• Confusing len() with capacity()

3. What will be the output of the following Rust code implementing a dynamic stack?

struct Stack {
    data: Vec<i32>,
}

impl Stack {
    fn new() -> Self {
        Stack { data: Vec::new() }
    }
    fn push(&mut self, val: i32) {
        self.data.push(val);
    }
    fn pop(&mut self) -> Option<i32> {
        self.data.pop()
    }
    fn print(&self) {
        for v in self.data.iter().rev() {
            print!("{} -> ", v);
        }
        println!("null");
    }
}

fn main() {
    let mut stack = Stack::new();
    stack.push(10);
    stack.push(20);
    stack.pop();
    stack.push(30);
    stack.print();
}

medium

A. 20 -> 10 -> null

B. 30 -> 20 -> null

C. 10 -> 30 -> null

D. 30 -> 10 -> null

Solution

1. Step 1: Trace push and pop operations

   Push 10: stack = [10]; Push 20: stack = [10, 20]; Pop: removes 20, stack = [10]; Push 30: stack = [10, 30]

2. Step 2: Understand print order

   Print iterates in reverse order: top is last element pushed, so prints 30 then 10, then null.

3. Final Answer:

   30 -> 10 -> null -> Option D

4. Quick Check:

   Stack top to bottom = 30, 10 [OK]

Hint: Pop removes last pushed; print reverses vector [OK]

Common Mistakes:

• Forgetting pop removes last element
• Printing in wrong order (not reversed)
• Confusing push order with print order

4. Identify the bug in this Rust code snippet for a dynamic stack push method:

fn push(&mut self, val: i32) {
    if self.data.len() == self.data.capacity() {
        self.data.reserve(self.data.len());
    }
    self.data.push(val);
}

medium

A. reserve argument should be capacity(), not len()

B. reserve should be called after push, not before

C. No bug; code works correctly

D. push should be called only if capacity is not full

Solution

1. Step 1: Check the resizing logic

   When self.data.len() == self.data.capacity(), the stack is full. reserve(self.data.len()) requests additional space for len() more elements. Since len() == capacity() at this point, it doubles the capacity.

2. Step 2: Verify correctness

   reserve(self.data.capacity()) would be equivalent here. The code safely resizes before push, preventing issues, and works correctly. (Rust Vec::push auto-resizes too, but manual is valid.)

3. Final Answer:

   No bug; code works correctly -> Option C

4. Quick Check:

   len() == capacity() when called, so doubles [OK]

Hint: When full, reserve(len()) == reserve(capacity()), doubles size [OK]

Common Mistakes:

• Reserving too little space (e.g., reserve(1))
• Calling reserve after push causing potential panic
• Assuming push auto-resizes without manual check

5. You want to implement a dynamic stack in Rust that shrinks its capacity when the number of elements falls below one-quarter of the capacity to save memory. Which approach correctly achieves this behavior?

hard

A. After pop, if len() < capacity() / 4, call data.shrink_to(len())

B. After pop, if len() < capacity() / 2, call data.reserve(len())

C. After pop, if len() < capacity() / 4, call data.reserve(capacity() / 2)

D. After pop, if len() < capacity() / 4, do nothing; shrinking is automatic

Solution

1. Step 1: Understand shrinking condition

   We want to shrink when elements are less than 25% of capacity to save memory.

2. Step 2: Use shrink_to method correctly

   Calling shrink_to(len()) reduces capacity to fit current elements, freeing unused space.

3. Step 3: Analyze options

   After pop, if len() < capacity() / 4, call data.shrink_to(len()) correctly checks condition and calls shrink_to(len()). Others either use wrong methods or do nothing.

4. Final Answer:

   After pop, if len() < capacity() / 4, call data.shrink_to(len()) -> Option A

5. Quick Check:

   Shrink when quarter full using shrink_to(len()) [OK]

Hint: Shrink vector capacity with shrink_to(len()) when quarter full [OK]

Common Mistakes:

• Using reserve instead of shrink_to for shrinking
• Shrinking at half capacity instead of quarter
• Assuming automatic shrinking without code