Why Dynamic Stack Using Resizable Array in DSA Rust?

The Big Idea

What if your stack could magically grow whenever you need more space, without you lifting a finger?

The Scenario

Imagine you have a stack of plates at home. You want to add more plates, but your shelf has a fixed size. When the shelf is full, you have to find a new shelf or manually move plates around to make space.

The Problem

Using a fixed-size array for a stack means you can run out of space quickly. You must guess the size in advance, and if you guess wrong, you waste space or run out. Manually resizing means copying all plates to a new shelf, which is slow and error-prone.

The Solution

A dynamic stack with a resizable array automatically grows when needed. It starts small and doubles its size when full, so you never run out of space. This makes adding plates easy and fast without manual moving.

Before vs After

✗ Before

let mut stack = [0; 5]; // fixed size
let mut top = 0;
// Can't push if full
if top < 5 {
  stack[top] = value;
  top += 1;
}

✓ After

let mut stack = Vec::new();
stack.push(value); // automatically resizes

What It Enables

You can build flexible stacks that grow as needed, making your programs handle more data smoothly without crashes or wasted space.

Real Life Example

Think of a web browser's back button history. It needs to remember pages you visited, but you don't know how many. A dynamic stack lets the browser store all pages you visit without limits.

Key Takeaways

Fixed-size stacks limit how much data you can store.

Dynamic stacks resize automatically to hold more items.

This makes your programs more flexible and reliable.

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