Why Stack Exists and What Problems It Solves in DSA Rust - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push an element onto the stack.

DSA Rust

stack.[1](5);

Drag options to blanks, or click blank then click option'

Apop

Bpush

Cpeek

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push removes an element instead of adding one.

Using peek only looks at the top element without adding.

The push method adds an element to the top of the stack.

2fill in blank

medium

Complete the code to remove the top element from the stack.

DSA Rust

let top = stack.[1]();

Drag options to blanks, or click blank then click option'

Apop

Bpush

Cpeek

Dinsert

Attempts:

3 left

💡 Hint

Common Mistakes

Using push adds an element instead of removing.

Using peek only views the top element without removing.

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

3fill in blank

hard

Fix the error in the code to check the top element without removing it.

DSA Rust

let top = stack.[1]().unwrap();

Drag options to blanks, or click blank then click option'

Apop

Bpush

Cremove

Dpeek

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop removes the element, which is not intended here.

Using push or remove are not valid for viewing the top element.

The peek method returns a reference to the top element without removing it.

4fill in blank

hard

Fill both blanks to create a stack and add an element to it.

DSA Rust

let mut stack = Vec::[1]();
stack.[2](10);

Drag options to blanks, or click blank then click option'

Anew

Bpop

Cpush

Dremove

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop or remove to create or add elements is incorrect.

Forgetting to make the stack mutable will cause errors.

Vec::new() creates a new empty stack (vector). push adds an element on top.

5fill in blank

hard

Fill all three blanks to pop the top element and print it if the stack is not empty.

DSA Rust

if let Some([1]) = stack.[2]() {
    println!("Top element: {}", [3]);
}

Drag options to blanks, or click blank then click option'

Avalue

Bpop

Dpeek

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek instead of pop will not remove the element.

Not handling the Option with if let can cause errors.

pop() removes the top element. if let Some(value) checks if stack is not empty. Then we print value.

Practice

1. Why do we use a stack data structure in programming?

easy

A. To keep track of function calls and return addresses

B. To store data in random order

C. To sort elements automatically

D. To store large amounts of data permanently

Solution

1. Step 1: Understand the role of stack in function calls

   Stacks help keep track of where each function should return after it finishes executing, by storing return addresses.

2. Step 2: Compare with other options

   Stacks do not sort data automatically, nor store data randomly or permanently; their main use is managing function calls.

3. Final Answer:

   To keep track of function calls and return addresses -> Option A

4. Quick Check:

   Stack = function call tracking [OK]

Hint: Stack remembers last function called first [OK]

Common Mistakes:

• Thinking stack sorts data automatically
• Confusing stack with permanent storage
• Assuming stack stores data randomly

2. Which of the following is the correct way to push an element onto a stack in Rust using a vector?

easy

A. stack.insert(element);

B. stack.add(element);

C. stack.append(element);

D. stack.push(element);

Solution

1. Step 1: Recall Rust vector methods for stack operations

   In Rust, a vector's push() method adds an element to the end, simulating stack push.

2. Step 2: Check other options for correctness

   Methods add(), insert(), and append() do not exist or do not behave like push() for stack usage.

3. Final Answer:

   stack.push(element); -> Option D

4. Quick Check:

   Use push() to add element to stack [OK]

Hint: Use push() to add on top of stack [OK]

Common Mistakes:

• Using add() which is not a Rust Vec method
• Confusing insert() which needs index
• Using append() which merges vectors

3. What will be the output of the following Rust code snippet?

let mut stack = Vec::new();
stack.push(10);
stack.push(20);
let top = stack.pop();
println!("{:?}", top);

medium

A. Some(20)

B. Some(10)

C. 20

D. None

Solution

1. Step 1: Trace stack push operations

   First 10 is pushed, then 20 is pushed. Stack now has [10, 20].

2. Step 2: Understand pop() behavior

   pop() removes and returns the last element, which is 20, wrapped in Some(20) because pop returns Option.

3. Final Answer:

   Some(20) -> Option A

4. Quick Check:

   pop() returns last element wrapped in Some [OK]

Hint: pop() returns Option with last pushed element [OK]

Common Mistakes:

• Expecting pop() to return raw value, not Option
• Confusing order of elements pushed
• Thinking pop() returns None when stack is not empty

4. Identify the error in this Rust code that uses a stack:

let mut stack = Vec::new();
let top = stack.pop().unwrap();
println!("{}", top);

medium

A. println! macro syntax is incorrect

B. Calling unwrap() on empty stack causes panic

C. stack should be immutable

D. pop() method does not exist on Vec

Solution

1. Step 1: Analyze pop() on empty vector

   pop() returns None if vector is empty. Calling unwrap() on None causes panic (runtime error).

2. Step 2: Check other options for validity

   pop() exists on Vec, stack is mutable, and println! syntax is correct.

3. Final Answer:

   Calling unwrap() on empty stack causes panic -> Option B

4. Quick Check:

   unwrap() on None panics [OK]

Hint: Never unwrap pop() without checking empty [OK]

Common Mistakes:

• Ignoring possibility of empty stack
• Thinking pop() is undefined on Vec
• Assuming println! syntax is wrong

5. You want to reverse a list of integers using a stack in Rust. Which approach correctly uses a stack to reverse the list?

let list = vec![1, 2, 3, 4];
let mut stack = Vec::new();
// Fill stack
for &item in &list {
    stack.push(item);
}
// Extract reversed list
let mut reversed = Vec::new();
while let Some(val) = stack.pop() {
    reversed.push(val);
}
println!("{:?}", reversed);

hard

A. Stack is not needed; list.reverse() is better

B. pop() should be replaced with peek() to reverse

C. This code correctly reverses the list using a stack

D. The code will panic because stack is empty

Solution

1. Step 1: Understand pushing all elements onto stack

   Each element from list is pushed onto stack in order: 1,2,3,4.

2. Step 2: Popping elements reverses order

   pop() removes last pushed element first, so reversed vector gets 4,3,2,1.

3. Step 3: Confirm no panic occurs

   while let Some(val) safely handles empty stack, no panic occurs.

4. Final Answer:

   This code correctly reverses the list using a stack -> Option C

5. Quick Check:

   Stack pop reverses order [OK]

Hint: Push all, then pop all to reverse [OK]

Common Mistakes:

• Replacing pop() with peek() which doesn't remove
• Assuming code panics on empty stack
• Ignoring that stack reverses order naturally