Implement Stack Using Queue in DSA Rust - Execution Trace

Concept Flow - Implement Stack Using Queue

Initialize empty queue

↓

Push operation: enqueue new element

↓

Rotate queue to move new element to front

↓

Pop operation: dequeue front element

↓

Top operation: peek front element

↓

Check if queue is empty

Stack operations are done by using a single queue where push adds element and then rotates the queue to keep the last pushed element at front, so pop and top operate on front.

Execution Sample

DSA Rust

queue = []
push(1)
push(2)
pop()
top()

This code pushes 1 and 2 onto the stack, pops the top element, then checks the top element.

Execution Table

Step	Operation	Queue State Before	Action	Queue State After	Visual State
1	Initialize queue	[]	Create empty queue	[]	[]
2	push(1)	[]	Enqueue 1, rotate 0 times	[1]	[1]
3	push(2)	[1]	Enqueue 2, rotate 1 time	[2, 1]	[2 -> 1]
4	pop()	[2, 1]	Dequeue front element (2)	[1]	[1]
5	top()	[1]	Peek front element	[1]	[1]
6	pop()	[1]	Dequeue front element (1)	[]	[]
7	pop()	[]	Queue empty, cannot pop	[]	[]

💡 Execution stops after attempting to pop from empty queue.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7
queue	[]	[1]	[2, 1]	[1]	[1]	[]	[]

Key Moments - 3 Insights

Why do we rotate the queue after pushing a new element?

Rotating the queue moves the newly added element to the front, so pop and top operations can access it directly as the stack's top. See step 3 in execution_table where after enqueueing 2, we rotate once to get [2, 1].

What happens if we pop when the queue is empty?

No element can be popped; the queue remains empty. Step 7 shows this case where pop is attempted on empty queue and no change occurs.

How does the queue represent the stack visually?

The front of the queue is the top of the stack. The visual state column shows the queue as elements connected with arrows, e.g., [2 -> 1] means 2 is top, then 1 below.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the queue state after push(2)?

A [1, 2]

B [2, 1]

C [2]

D [1]

💡 Hint

Check the 'Queue State After' column at step 3 in execution_table.

At which step does the queue become empty after popping all elements?

A Step 4

B Step 5

C Step 6

D Step 7

💡 Hint

Look at the 'Queue State After' column to see when queue becomes [].

If we skip rotating the queue after push, what would be the top element after pushing 2?

A 1

B 2

C Empty

D Error

💡 Hint

Without rotation, the new element stays at the back, so front remains the old element. See explanation in key_moments about rotation.

Concept Snapshot

Stack using one queue:
- Push: enqueue element, then rotate queue to front
- Pop: dequeue front element
- Top: peek front element
- Empty: check if queue is empty
Rotation keeps last pushed element at front for stack behavior.

Full Transcript

This concept shows how to implement a stack using a single queue. We start with an empty queue. When pushing, we add the new element to the back and then rotate the queue so this new element moves to the front. This way, the front of the queue always represents the top of the stack. Popping removes the front element, and top peeks at the front. If the queue is empty, pop does nothing. The execution table traces each step showing queue states and actions. Key moments clarify why rotation is needed and what happens when popping empty. The visual quiz tests understanding of queue states and rotation effects. The snapshot summarizes the method in short lines.

Practice

1. What is the main idea behind implementing a stack using a queue?

easy

A. Use one or two queues to simulate stack operations like push and pop

B. Use a queue to sort elements in ascending order

C. Use a stack to implement queue operations

D. Use two stacks to implement a queue

Solution

1. Step 1: Understand stack and queue operations

   A stack follows Last-In-First-Out (LIFO), while a queue follows First-In-First-Out (FIFO).

2. Step 2: Simulate stack using queue(s)

   By using one or two queues and carefully managing element order, we can mimic stack push and pop operations.

3. Final Answer:

   Use one or two queues to simulate stack operations like push and pop -> Option A

4. Quick Check:

   Stack via queue = simulate push/pop with queues [OK]

Hint: Remember stack is LIFO, queue is FIFO; simulate LIFO with queue(s) [OK]

Common Mistakes:

• Confusing stack and queue order
• Trying to sort elements instead of simulating stack
• Using stack to implement queue instead

2. Which Rust code snippet correctly declares a queue using VecDeque for implementing a stack?

easy

A. let mut queue = Vec::with_capacity(10);

B. let queue = Vec<i32>::new();

C. let mut queue: VecDeque<i32> = VecDeque::new();

D. let queue: VecDeque = Vec::new();

Solution

1. Step 1: Identify correct queue type in Rust

   VecDeque is the standard double-ended queue in Rust, suitable for queue operations.

2. Step 2: Check correct syntax for mutable queue declaration

   "let mut queue: VecDeque<i32> = VecDeque::new();" correctly declares a mutable queue of i32.

3. Final Answer:

   let mut queue: VecDeque<i32> = VecDeque::new(); -> Option C

4. Quick Check:

   VecDeque mutable queue declaration = let mut queue: VecDeque<i32> = VecDeque::new(); [OK]

Hint: Use VecDeque with mut keyword for queue in Rust [OK]

Common Mistakes:

• Using Vec instead of VecDeque for queue
• Missing mut keyword for mutable queue
• Wrong type annotations

3. Given this Rust code implementing push for stack using one queue, what is the queue state after pushing 1, then 2?

use std::collections::VecDeque;

struct Stack {
    queue: VecDeque<i32>,
}

impl Stack {
    fn new() -> Self {
        Stack { queue: VecDeque::new() }
    }

    fn push(&mut self, x: i32) {
        self.queue.push_back(x);
        let size = self.queue.len();
        for _ in 0..size - 1 {
            if let Some(val) = self.queue.pop_front() {
                self.queue.push_back(val);
            }
        }
    }
}

fn main() {
    let mut stack = Stack::new();
    stack.push(1);
    stack.push(2);
    println!("{:?}", stack.queue);
}

medium

A. [2, 1]

B. [1, 2]

C. [1]

D. [2]

Solution

1. Step 1: Trace push(1) operation

   Push 1: queue = [1]. No rotation since size=1.

2. Step 2: Trace push(2) operation

   Push 2: queue = [1, 2]. Rotate size-1=1 times: pop_front 1, push_back 1 -> queue = [2, 1].

3. Final Answer:

   [2, 1] -> Option A

4. Quick Check:

   After push 1,2 queue = [2,1] [OK]

Hint: After push, rotate queue to front to simulate stack top [OK]

Common Mistakes:

• Not rotating queue after push
• Confusing push_back and pop_front order
• Assuming queue keeps insertion order

4. Identify the bug in this pop method for stack implemented using one queue:

fn pop(&mut self) -> Option<i32> {
    self.queue.pop_back()
}

medium

A. pop_back() is invalid for VecDeque

B. pop_back() removes from the wrong end; should use pop_front()

C. pop_back() is correct; no bug

D. pop_back() requires mutable borrow error

Solution

1. Step 1: Understand queue ends in VecDeque

   VecDeque supports pop_front() and pop_back(), but our stack top is at front after push rotation.

2. Step 2: Check which end holds stack top

   After push rotation, stack top is at front, so pop_front() removes top, not pop_back().

3. Final Answer:

   pop_back() removes from the wrong end; should use pop_front() -> Option B

4. Quick Check:

   Stack top at front, pop_front() needed [OK]

Hint: Pop from front after rotation to get stack top [OK]

Common Mistakes:

• Using pop_back() instead of pop_front()
• Not rotating queue after push
• Confusing queue ends

5. How would you implement the top method for a stack using a single queue in Rust, assuming push rotates the queue so the top is at the front?

hard

A. Return the last element using queue.back().cloned()

B. Remove and return the last element using queue.pop_back()

C. Remove and return the first element using queue.pop_front()

D. Return the element at the front of the queue without removing it using queue.front().cloned()

Solution

1. Step 1: Recall queue rotation after push

   After push, queue is rotated so top element is at front.

2. Step 2: Access top without removing

   Use queue.front().cloned() to get top element without popping.

3. Final Answer:

   Return the element at the front of the queue without removing it using queue.front().cloned() -> Option D

4. Quick Check:

   Top element at front, use front().cloned() [OK]

Hint: Top is front element after rotation; use front().cloned() [OK]

Common Mistakes:

• Using pop_front() which removes element
• Using back() which is bottom of stack
• Removing element instead of just viewing