Peek Top Element of Stack in DSA Rust - Build from Scratch

Peek Top Element of Stack

📖 Scenario: You are building a simple program to manage a stack of books. You want to see which book is currently on top without removing it.

🎯 Goal: Create a stack using a vector, add some books, then peek at the top book without removing it.

📋 What You'll Learn

Create a vector called stack with exact book titles

Add a variable called top_book to hold the peeked book

Use the last() method to peek at the top book

Print the top_book value

💡 Why This Matters

🌍 Real World

Stacks are used in many real-world applications like undo features in text editors, browser history, and managing function calls.

💼 Career

Understanding how to peek at the top of a stack is important for software developers working with data structures, algorithms, and system design.

1

Create the stack with books

Create a vector called stack with these exact book titles as strings: "The Hobbit", "1984", "Brave New World"

DSA Rust

# Create a vector called stack with the given book titles
# Your code here

Hint

Use vec![...] to create a vector with the exact strings inside.

2

Add a variable to hold the top book

Add a variable called top_book that uses stack.last() to peek at the top book without removing it

DSA Rust

fn main() {
    let stack = vec!["The Hobbit", "1984", "Brave New World"];
    // Add a variable top_book to peek at the last element
    // Your code here
}

Hint

Use stack.last() to get a reference to the last element in the vector.

3

Handle the peeked value safely

Use a match statement on top_book to get the book title or print "Stack is empty" if there is no top book

DSA Rust

fn main() {
    let stack = vec!["The Hobbit", "1984", "Brave New World"];
    let top_book = stack.last();
    // Use match on top_book to get the book or print "Stack is empty"
    // Your code here
}

Hint

Use match to check if top_book is Some or None.

4

Print the top book

Print the top book using println!("Top book: {}", book) inside the Some arm of the match statement

DSA Rust

fn main() {
    let stack = vec!["The Hobbit", "1984", "Brave New World"];
    let top_book = stack.last();
    match top_book {
        Some(book) => {
            // Print the top book here
            // Your code here
        },
        None => println!("Stack is empty"),
    }
}

Hint

The program should print exactly: Top book: Brave New World

Practice

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

easy

A. Returns the top element without removing it

B. Removes the top element and returns it

C. Adds a new element to the top

D. Checks if the stack is empty

Solution

1. Step 1: Understand stack operations

   A stack stores elements in a last-in, first-out order. Common operations include push, pop, and peek.

2. Step 2: Define peek operation

   Peek returns the top element without removing it, allowing you to see the current top value safely.

3. Final Answer:

   Returns the top element without removing it -> Option A

4. Quick Check:

   Peek = top element without removal [OK]

Hint: Peek shows top without changing stack [OK]

Common Mistakes:

• Confusing peek with pop operation
• Thinking peek adds elements
• Assuming peek checks emptiness

2. Which of the following Rust code snippets correctly peeks the top element of a stack implemented as a Vec<i32> named stack?

easy

A. let top = stack.last();

B. let top = stack[0];

C. let top = stack.pop();

D. let top = stack.first();

Solution

1. Step 1: Understand Rust Vec methods for stack

   In Rust, Vec can be used as a stack. pop() removes and returns the last element, last() returns a reference to the last element without removing it.

2. Step 2: Identify correct peek method

   Peek means to see the top element without removing it, so stack.last() is correct. stack.pop() removes it, stack[0] accesses the first element, and stack.first() returns the first element, not the top.

3. Final Answer:

   let top = stack.last(); -> Option A

4. Quick Check:

   Peek = stack.last() [OK]

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

Common Mistakes:

• Using pop() which removes element
• Accessing first() instead of last()
• Using index 0 which is bottom of stack

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

let mut stack = vec![10, 20, 30];
if let Some(top) = stack.last() {
    println!("{}", top);
} else {
    println!("Stack is empty");
}

medium

A. 20

B. 30

C. 10

D. Stack is empty

Solution

1. Step 1: Understand stack contents and last()

   The stack vector contains [10, 20, 30]. The last element is 30.

2. Step 2: Trace the if let statement

   The stack.last() returns Some(&30), so the if block prints 30.

3. Final Answer:

   30 -> Option B

4. Quick Check:

   stack.last() = 30 [OK]

Hint: stack.last() returns top element reference [OK]

Common Mistakes:

• Printing bottom element instead of top
• Confusing last() with pop() output
• Assuming stack is empty

4. Identify the error in the following Rust code that tries to peek the top element of a stack and fix it:

let stack = vec![1, 2, 3];
let top = stack.pop().unwrap();
println!("Top element: {}", top);

medium

A. No error, code works correctly

B. pop() removes element; use last() to peek instead

C. unwrap() causes panic; use expect() instead

D. stack should be mutable to use pop()

Solution

1. Step 1: Check mutability for pop()

   pop() removes the last element, so the vector must be mutable. Here, stack is immutable.

2. Step 2: Fix by making stack mutable

   Change let stack = vec![1, 2, 3]; to let mut stack = vec![1, 2, 3]; to allow pop().

3. Final Answer:

   stack should be mutable to use pop() -> Option D

4. Quick Check:

   pop() needs mutable stack [OK]

Hint: pop() requires mutable stack variable [OK]

Common Mistakes:

• Ignoring mut keyword error
• Confusing pop() with peek operation
• Assuming unwrap() always safe

5. Given a stack implemented as Vec<i32>, write a Rust function peek_top that returns the top element as Option<i32> without removing it. What is the correct implementation?

hard

A. fn peek_top(stack: &mut Vec<i32>) -> Option<i32> { stack.pop() }

B. fn peek_top(stack: &Vec<i32>) -> i32 { *stack.last().unwrap() }

C. fn peek_top(stack: &Vec<i32>) -> Option<i32> { stack.last().copied() }

D. fn peek_top(stack: Vec<i32>) -> Option<i32> { stack.first().copied() }

Solution

1. Step 1: Understand function requirements

   The function must peek without removing, so it should borrow stack immutably and return Option<i32>.

2. Step 2: Analyze options

   fn peek_top(stack: &Vec<i32>) -> Option<i32> { stack.last().copied() } uses stack.last().copied() which returns Option<i32> by copying the reference. fn peek_top(stack: &mut Vec<i32>) -> Option<i32> { stack.pop() } removes element (pop), fn peek_top(stack: &Vec<i32>) -> i32 { *stack.last().unwrap() } unwraps and returns i32 (not Option), fn peek_top(stack: Vec<i32>) -> Option<i32> { stack.first().copied() } uses first() which is bottom element.

3. Final Answer:

   fn peek_top(stack: &Vec<i32>) -> Option<i32> { stack.last().copied() } -> Option C

4. Quick Check:

   Peek returns Option with copied last element [OK]

Hint: Use last().copied() for safe peek returning Option<i32> [OK]

Common Mistakes:

• Using pop() which removes element
• Returning i32 instead of Option<i32>
• Accessing first() instead of last()