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DSA Rustprogramming~10 mins

Dynamic Stack Using Resizable Array in DSA Rust - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to create a new empty stack.

DSA Rust
let mut stack: Vec<i32> = Vec::[1]();
Drag options to blanks, or click blank then click option'
Adefault
Bwith_capacity
Cfrom
Dnew
Attempts:
3 left
💡 Hint
Common Mistakes
Using Vec::with_capacity() without specifying capacity.
Using Vec::from() which requires an initial array or slice.
2fill in blank
medium

Complete the code to push an element onto the stack.

DSA Rust
stack.[1](10);
Drag options to blanks, or click blank then click option'
Ainsert
Bappend
Cpush
Dextend
Attempts:
3 left
💡 Hint
Common Mistakes
Using insert() which requires an index.
Using append() which adds elements from another vector.
3fill in blank
hard

Fix the error in popping an element from the stack safely.

DSA Rust
let top = stack.[1]().unwrap_or(-1);
Drag options to blanks, or click blank then click option'
Apop
Bget
Clast
Dremove
Attempts:
3 left
💡 Hint
Common Mistakes
Using last() which only returns a reference without removing.
Using remove() which requires an index.
4fill in blank
hard

Fill both blanks to check if the stack is empty and print a message.

DSA Rust
if stack.[1]() [2] 0 {
    println!("Stack is empty");
}
Drag options to blanks, or click blank then click option'
Alen
B==
C!=
Dis_empty
Attempts:
3 left
💡 Hint
Common Mistakes
Using is_empty() without comparison.
Using != instead of == causing wrong condition.
5fill in blank
hard

Fill all three blanks to implement a function that returns the top element without removing it.

DSA Rust
fn peek(stack: &Vec<i32>) -> Option<i32> {
    stack.[1]().cloned().[2](|| [3])
}
Drag options to blanks, or click blank then click option'
Alast
Bunwrap_or
CNone
Dunwrap
Attempts:
3 left
💡 Hint
Common Mistakes
Using unwrap() which panics if stack is empty.
Using pop() which removes the element.

Practice

(1/5)
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
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10 questions~5 mineasy hard