Pop Operation on Stack in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after popping from the stack?

Consider a stack implemented as a vector in Rust. What will be the printed stack after popping the top element?

DSA Rust

let mut stack = vec![10, 20, 30, 40];
stack.pop();
println!("{:?}", stack);

A []

B [10, 20, 30, 40]

C [20, 30, 40]

D [10, 20, 30]

💡 Hint

Popping removes the last element added to the stack.

The pop operation removes the last element (40) from the stack, leaving [10, 20, 30].

❓ Predict Output

intermediate

What happens when popping from an empty stack?

What is the output of the following Rust code when popping from an empty stack?

DSA Rust

let mut stack: Vec<i32> = Vec::new();
let popped = stack.pop();
println!("{:?}", popped);

A None

B Some(-1)

C Some(0)

D panic at runtime

💡 Hint

pop() returns an Option type in Rust.

Popping from an empty vector returns None, indicating no value was removed.

🔧 Debug

advanced

Identify the error in this pop operation code

What error will this Rust code produce when trying to pop from the stack?

DSA Rust

let mut stack = vec![1, 2, 3];
let value = stack.pop().unwrap();
println!("Popped: {}", value);
let value2 = stack.pop().unwrap();
let value3 = stack.pop().unwrap();
let value4 = stack.pop().unwrap();

A TypeError because pop returns an integer

B SyntaxError due to missing semicolon

C panic at runtime due to unwrap on None

D No error, prints all popped values

💡 Hint

unwrap() on None causes panic.

The fourth pop tries to unwrap None because the stack is empty, causing a runtime panic.

🧠 Conceptual

advanced

What is the time complexity of pop operation on a stack implemented with a vector?

Choose the correct time complexity for the pop operation on a stack implemented using a Rust vector.

A O(1) - Constant time

B O(n^2) - Quadratic time

C O(log n) - Logarithmic time

D O(n) - Linear time

💡 Hint

Pop removes the last element directly.

Pop removes the last element in constant time without shifting elements.

🚀 Application

expert

After a sequence of push and pop operations, what is the final stack?

Given the following Rust code, what is the final state of the stack printed?

DSA Rust

let mut stack = Vec::new();
stack.push(5);
stack.push(10);
stack.push(15);
stack.pop();
stack.push(20);
stack.pop();
stack.pop();
println!("{:?}", stack);

A []

B [5]

C [5, 10]

D [10]

💡 Hint

Track each push and pop step carefully.

Push 5,10,15 → stack: [5,10,15] Pop → removes 15 → [5,10] Push 20 → [5,10,20] Pop → removes 20 → [5,10] Pop → removes 10 → [5] Final stack is [5].

Practice

1. What does the pop operation do in a stack data structure?

easy

A. Adds a new element to the bottom of the stack

B. Removes and returns the top element of the stack

C. Returns the bottom element without removing it

D. Clears all elements from the stack

Solution

1. Step 1: Understand stack behavior

   A stack follows Last In First Out (LIFO) principle, so the last added element is on top.

2. Step 2: Define pop operation

   The pop operation removes and returns the top element, which is the last inserted item.

3. Final Answer:

   Removes and returns the top element of the stack -> Option B

4. Quick Check:

   Pop = remove top element [OK]

Hint: Pop always removes the last added item [OK]

Common Mistakes:

• Thinking pop removes from bottom
• Confusing pop with push
• Assuming pop returns without removing

2. Which of the following is the correct Rust syntax to pop an element from a stack implemented as a Vec<T> named stack?

easy

A. let x = stack.pop_front();

B. let x = stack.remove();

C. let x = stack.pop();

D. let x = stack.delete();

Solution

1. Step 1: Recall Rust Vec methods

   Rust's Vec has a pop() method that removes and returns the last element.

2. Step 2: Check other options

   remove() requires an index, pop_front() is for other collections, delete() does not exist.

3. Final Answer:

   let x = stack.pop(); -> Option C

4. Quick Check:

   Vec pop method = stack.pop() [OK]

Hint: Use stack.pop() to remove last element in Rust Vec [OK]

Common Mistakes:

• Using pop_front() which is not in Vec
• Calling remove() without index
• Using non-existent delete() method

3. What is the output of the following Rust code?

let mut stack = vec![10, 20, 30];
let popped = stack.pop();
println!("{:?}", popped);
println!("{:?}", stack);

medium

A. Some(30)
[10, 20]

B. 30
[10, 20, 30]

C. None
[10, 20, 30]

D. Some(20)
[10, 30]

Solution

1. Step 1: Understand pop() return type

   pop() returns an Option containing the last element if present, here Some(30).

2. Step 2: Check stack after pop

   After popping 30, stack becomes [10, 20].

3. Final Answer:

   Some(30) [10, 20] -> Option A

4. Quick Check:

   pop returns Some(last element), stack shrinks [OK]

Hint: pop() returns Some(value) if not empty [OK]

Common Mistakes:

• Expecting pop() to return value directly, not Option
• Not updating stack after pop
• Confusing popped element with stack contents

4. Identify the error in this Rust code that tries to pop an element from an empty stack:

let mut stack: Vec<i32> = Vec::new();
let x = stack.pop().unwrap();
println!("{}", x);

medium

A. pop() method does not exist on Vec

B. println! macro syntax is incorrect

C. stack is not mutable

D. Calling unwrap() on None causes panic

Solution

1. Step 1: Check pop() on empty Vec

   pop() returns None if the Vec is empty.

2. Step 2: Understand unwrap() behavior

   Calling unwrap() on None causes a runtime panic (crash).

3. Final Answer:

   Calling unwrap() on None causes panic -> Option D

4. Quick Check:

   unwrap on None = panic [OK]

Hint: Avoid unwrap() on pop() without checking Some [OK]

Common Mistakes:

• Assuming pop() always returns Some
• Ignoring mut keyword needed for pop
• Misreading println! syntax

5. Given a stack implemented as Vec<i32>, how can you safely pop all elements and collect them into a new vector in reverse order without panicking?

hard

A. Use a loop with while let Some(x) = stack.pop() and push x into new vector

B. Use stack.pop().unwrap() repeatedly until empty

C. Use stack.remove(0) repeatedly to pop elements

D. Use stack.pop_front() to remove elements safely

Solution

1. Step 1: Avoid unwrap() to prevent panic

   Using unwrap() on pop() can panic if stack is empty, so avoid it.

2. Step 2: Use while let to safely pop

   Using while let Some(x) = stack.pop() safely extracts elements until empty.

3. Step 3: Push popped elements into new vector

   Each popped element x is pushed into a new vector, reversing order.

4. Final Answer:

   Use a loop with while let Some(x) = stack.pop() and push x into new vector -> Option A

5. Quick Check:

   Safe pop loop = while let Some(x) = pop() [OK]

Hint: Use while let Some(x) = pop() to avoid panic [OK]

Common Mistakes:

• Using unwrap() causing panic on empty stack
• Using remove(0) which is inefficient
• Calling pop_front() which does not exist on Vec