Queue Using Two Stacks in DSA Rust - Build from Scratch

Queue Using Two Stacks

📖 Scenario: Imagine you are managing a line of customers waiting for service. You want to build a queue that follows the first-come, first-served rule. However, you only have two stacks to help you. Your task is to create a queue using these two stacks.

🎯 Goal: Build a queue data structure using two stacks in Rust. You will implement enqueue and dequeue operations using these stacks and then print the queue state after some operations.

📋 What You'll Learn

Create two stacks using Vec<i32> named stack_in and stack_out

Create a function enqueue that adds elements to stack_in

Create a function dequeue that removes elements from stack_out, transferring elements from stack_in if needed

Print the final state of the queue after some enqueue and dequeue operations

💡 Why This Matters

🌍 Real World

Queues are used in many real-world systems like customer service lines, task scheduling, and buffering data streams. Using two stacks to build a queue is a common interview problem that shows how to use simple data structures to build more complex ones.

💼 Career

Understanding how to implement queues with stacks helps in problem-solving and coding interviews. It also improves your grasp of data structure manipulation and algorithm design.

1

Create Two Stacks

Create two empty stacks called stack_in and stack_out using Vec<i32>.

DSA Rust

# Create two empty stacks named stack_in and stack_out
// Your code here

Hint

Use Vec::new() to create empty stacks.

2

Define Enqueue Function

Define a function called enqueue that takes a mutable reference to stack_in and an i32 value, then pushes the value onto stack_in.

DSA Rust

let mut stack_in: Vec = Vec::new();
let mut stack_out: Vec = Vec::new();

// Define enqueue function here
// Your code here

Hint

Use push method on stack_in to add the value.

3

Define Dequeue Function

Define a function called dequeue that takes mutable references to stack_in and stack_out. If stack_out is empty, move all elements from stack_in to stack_out. Then pop and return the top element from stack_out. Return None if both stacks are empty.

DSA Rust

let mut stack_in: Vec = Vec::new();
let mut stack_out: Vec = Vec::new();

fn enqueue(stack_in: &mut Vec, value: i32) {
    stack_in.push(value);
}

// Define dequeue function here
// Your code here

Hint

Move elements from stack_in to stack_out only if stack_out is empty, then pop from stack_out.

4

Use Queue and Print Final State

Use the enqueue function to add 10, 20, and 30 to the queue. Then call dequeue once. Finally, print the contents of stack_out followed by the reversed contents of stack_in to show the queue state from front to back.

DSA Rust

let mut stack_in: Vec = Vec::new();
let mut stack_out: Vec = Vec::new();

fn enqueue(stack_in: &mut Vec, value: i32) {
    stack_in.push(value);
}

fn dequeue(stack_in: &mut Vec, stack_out: &mut Vec) -> Option {
    if stack_out.is_empty() {
        while let Some(val) = stack_in.pop() {
            stack_out.push(val);
        }
    }
    stack_out.pop()
}

// Enqueue 10, 20, 30
// Dequeue once
// Print queue state from front to back
// Your code here

Hint

After dequeue, the queue should have 20 and 30 in order. Print the combined stacks to show this.

Practice

1. What is the main idea behind implementing a Queue Using Two Stacks?

easy

A. Use one stack for enqueue and another for dequeue operations

B. Use two stacks to perform only enqueue operations

C. Use two stacks to store elements in sorted order

D. Use two stacks to reverse the order twice to maintain queue order

Solution

1. Step 1: Understand queue order

   A queue follows First-In-First-Out (FIFO) order, meaning the first element added is the first removed.

2. Step 2: Use two stacks to reverse order twice

   One stack reverses the order when pushing elements, and the second stack reverses it back when popping, preserving FIFO.

3. Final Answer:

   Use two stacks to reverse the order twice to maintain queue order -> Option D

4. Quick Check:

   Two stacks reverse order twice = queue order [OK]

Hint: Two stacks reverse order twice to simulate queue [OK]

Common Mistakes:

• Thinking both stacks store elements in sorted order
• Using stacks only for enqueue or dequeue
• Confusing stack order with queue order

2. Which of the following Rust code snippets correctly declares two stacks for implementing a queue?

easy

A. let mut stack1: Vec<i32> = Vec::new(); let mut stack2: Vec<i32> = Vec::new();

B. let stack1 = Vec<i32>; let stack2 = Vec<i32>;

C. let mut stack1 = Vec::new<i32>(); let mut stack2 = Vec::new<i32>();

D. let stack1: Vec = Vec::new(); let stack2: Vec = Vec::new();

Solution

1. Step 1: Check correct Vec declaration syntax in Rust

   In Rust, to declare a mutable vector of i32, use: let mut stack: Vec<i32> = Vec::new();

2. Step 2: Compare options with correct syntax

   let mut stack1: Vec<i32> = Vec::new(); let mut stack2: Vec<i32> = Vec::new(); matches the correct syntax exactly; others have syntax errors or missing mutability.

3. Final Answer:

   let mut stack1: Vec<i32> = Vec::new(); let mut stack2: Vec<i32> = Vec::new(); -> Option A

4. Quick Check:

   Correct Rust Vec declaration = let mut stack1: Vec<i32> = Vec::new(); let mut stack2: Vec<i32> = Vec::new(); [OK]

Hint: Use let mut stack: Vec<i32> = Vec::new() for mutable stacks [OK]

Common Mistakes:

• Omitting mut keyword for mutable vectors
• Incorrect generic syntax like Vec<i32> without parentheses
• Using Vec::new<i32>() which is invalid

3. What will be the output of the following Rust code implementing a queue using two stacks?

struct Queue {
    stack1: Vec<i32>,
    stack2: Vec<i32>,
}

impl Queue {
    fn new() -> Self {
        Queue { stack1: Vec::new(), stack2: Vec::new() }
    }

    fn enqueue(&mut self, val: i32) {
        self.stack1.push(val);
    }

    fn dequeue(&mut self) -> Option<i32> {
        if self.stack2.is_empty() {
            while let Some(x) = self.stack1.pop() {
                self.stack2.push(x);
            }
        }
        self.stack2.pop()
    }
}

fn main() {
    let mut q = Queue::new();
    q.enqueue(10);
    q.enqueue(20);
    q.enqueue(30);
    println!("{:?}", q.dequeue());
    println!("{:?}", q.dequeue());
    q.enqueue(40);
    println!("{:?}", q.dequeue());
    println!("{:?}", q.dequeue());
}

medium

A. Some(10) Some(20) Some(30) Some(40)

B. Some(30) Some(20) Some(10) Some(40)

C. Some(10) Some(20) Some(40) None

D. None None None None

Solution

1. Step 1: Trace enqueue operations

   Enqueue 10, 20, 30 pushes these onto stack1 in order: [10, 20, 30]

2. Step 2: Trace dequeue operations

   First dequeue: stack2 empty, pop all from stack1 to stack2: stack2 = [30, 20, 10]. Pop from stack2 returns 10.
   Second dequeue: pop from stack2 returns 20.
   Enqueue 40: push 40 to stack1.
   Third dequeue: pop from stack2 returns 30.
   Fourth dequeue: stack2 empty, move 40 from stack1 to stack2, pop returns 40.

3. Final Answer:

   Some(10) Some(20) Some(30) Some(40) -> Option A

4. Quick Check:

   Queue order preserved by two stacks = Some(10) Some(20) Some(30) Some(40) [OK]

Hint: Dequeue empties stack2 by reversing stack1 first [OK]

Common Mistakes:

• Assuming stack2 pops in same order as stack1
• Not moving elements from stack1 to stack2 when stack2 is empty
• Confusing order of elements after transfer

4. Identify the bug in this Rust code snippet for dequeue operation in a queue using two stacks:

fn dequeue(&mut self) -> Option<i32> {
    while let Some(x) = self.stack1.pop() {
        self.stack2.push(x);
    }
    self.stack2.pop()
}

medium

A. It uses pop() on stack1 instead of peek()

B. It moves elements from stack1 to stack2 every time, even if stack2 is not empty

C. It does not pop from stack2 after moving elements

D. It returns None always because stack2 is never filled

Solution

1. Step 1: Analyze the dequeue logic

   The code moves all elements from stack1 to stack2 every dequeue call, regardless of stack2's state.

2. Step 2: Identify correct behavior

   To maintain queue order efficiently, elements should be moved only if stack2 is empty. Otherwise, stack2's elements should be popped directly.

3. Final Answer:

   It moves elements from stack1 to stack2 every time, even if stack2 is not empty -> Option B

4. Quick Check:

   Move only if stack2 empty = It moves elements from stack1 to stack2 every time, even if stack2 is not empty [OK]

Hint: Move elements only if stack2 is empty during dequeue [OK]

Common Mistakes:

• Moving elements every dequeue wastes time
• Ignoring stack2's current elements
• Not popping from stack2 after transfer

5. You want to implement a peek function for the queue using two stacks in Rust that returns the front element without removing it. Which approach correctly achieves this?

hard

A. Return the first element of stack1 without any stack transfers

B. Return the last element of stack1 directly without moving elements

C. If stack2 is empty, move all elements from stack1 to stack2, then return the last element of stack2 without popping

D. Pop an element from stack2 and push it back immediately to peek

Solution

1. Step 1: Understand peek operation in queue using two stacks

   Peek should return the front element without removing it. The front is at the bottom of stack1 or top of stack2 after transfer.

2. Step 2: Correct approach for peek

   If stack2 is empty, move all elements from stack1 to stack2 to reverse order. Then peek the last element of stack2 (top) without popping.

3. Final Answer:

   If stack2 is empty, move all elements from stack1 to stack2, then return the last element of stack2 without popping -> Option C

4. Quick Check:

   Peek uses stack2 top after transfer if empty = If stack2 is empty, move all elements from stack1 to stack2, then return the last element of stack2 without popping [OK]

Hint: Peek top of stack2 after transfer if empty [OK]

Common Mistakes:

• Peeking stack1 directly ignores queue order
• Popping to peek then pushing back wastes time
• Ignoring stack2 state before peeking