Queue Using Two Stacks in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind implementing a queue using two stacks?

Use one stack to handle incoming elements (enqueue) and another stack to reverse the order for outgoing elements (dequeue), simulating the FIFO behavior of a queue.

beginner

Why do we need two stacks to implement a queue instead of just one?

A single stack follows LIFO (last-in, first-out), but a queue needs FIFO (first-in, first-out). Two stacks help reverse the order of elements to achieve FIFO behavior.

intermediate

In the two-stack queue, what happens during the dequeue operation if the output stack is empty?

All elements from the input stack are popped and pushed into the output stack, reversing their order, so the oldest element is on top and can be dequeued.

intermediate

What is the time complexity of enqueue and dequeue operations in a queue implemented with two stacks?

Enqueue is O(1) because it just pushes to the input stack. Dequeue is amortized O(1) because elements are moved between stacks only when needed.

beginner

Show a simple Rust function signature for enqueue and dequeue in a two-stack queue implementation.

fn enqueue(&mut self, value: T); fn dequeue(&mut self) -> Option<T>;

What data structure is used internally to reverse the order of elements in a two-stack queue?

A Queue

B Array

C Linked List

D Stack

Two stacks are used; one stack reverses the order of elements to simulate queue behavior.

If the output stack is not empty, what does the dequeue operation do?

A Returns the top element from output stack

B Clears both stacks

C Pushes new element to input stack

D Moves elements from input stack to output stack

If output stack has elements, dequeue returns the top element directly.

What is the worst-case time complexity of a single dequeue operation in a two-stack queue?

A O(n)

B O(1)

C O(log n)

D O(n log n)

In worst case, all elements move from input to output stack, which is O(n).

Which operation is always O(1) in a two-stack queue?

A Dequeue

B Enqueue

C Moving elements between stacks

D Checking if queue is empty

Enqueue just pushes to input stack, which is O(1).

What happens if you dequeue from an empty two-stack queue?

A Returns last enqueued element

B Throws an error

C Returns None or equivalent

D Blocks until element is available

Dequeue returns None (or equivalent) when queue is empty.

Explain how two stacks can be used to implement a queue. Describe the enqueue and dequeue steps.

Think about how stacks reverse order and how that helps simulate a queue.

Describe the time complexity of enqueue and dequeue operations in a queue implemented with two stacks and why.

Consider how often elements move between stacks.

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