Dynamic Stack Using Resizable Array in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is a dynamic stack using a resizable array?

A dynamic stack using a resizable array is a stack data structure that grows or shrinks its underlying array automatically when elements are pushed or popped, allowing flexible size without fixed limits.

beginner

Why do we resize the array in a dynamic stack?

We resize the array to handle more elements when the stack is full or to save memory when many elements are removed, ensuring efficient use of space and avoiding overflow errors.

beginner

In Rust, which method is commonly used to add an element to the end of a vector?

The method is push(), which appends an element to the end of the vector, increasing its length by one.

beginner

What happens when you pop an element from an empty dynamic stack?

Popping from an empty dynamic stack returns None or an error, depending on implementation, because there are no elements to remove.

intermediate

How does doubling the array size during resizing help performance?

Doubling the array size reduces the number of times resizing happens, making push operations mostly fast (amortized constant time), balancing speed and memory use.

What is the main advantage of using a dynamic stack over a fixed-size stack?

A It never needs resizing.

B It uses less memory always.

C It can grow or shrink as needed.

D It only stores integers.

A dynamic stack can grow or shrink its storage array as elements are added or removed, unlike a fixed-size stack.

In Rust, which data structure is typically used to implement a dynamic stack?

A Vec (vector)

B LinkedList

C HashMap

D Array fixed size

Vec is a resizable array in Rust and is commonly used to implement dynamic stacks.

What happens internally when the stack's underlying array is full and a new element is pushed?

A The array is resized to a larger capacity.

B The new element is discarded.

C The stack throws an error immediately.

D The array shrinks.

When full, the array is resized (usually doubled) to accommodate more elements.

Which operation removes the top element from the stack?

A push

B pop

C peek

D insert

The pop operation removes and returns the top element from the stack.

Why is doubling the array size during resizing preferred over increasing by one element?

A It causes errors.

B It wastes more memory.

C It slows down push operations.

D It reduces the total number of resizes needed.

Doubling reduces how often resizing happens, improving overall performance.

Explain how a dynamic stack using a resizable array works, including how it handles pushing and popping elements.

Think about what happens when the array is full or empty.

Describe why resizing the underlying array in a dynamic stack is important and how it affects performance.

Consider both adding and removing elements.

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