Queue Using Two Stacks in DSA Javascript - Build from Scratch

Queue Using Two Stacks

📖 Scenario: Imagine you are managing a line of customers waiting for service. You want to keep track of who comes first and who leaves first, like a queue. But you only have two stacks (like piles of plates) to help you. Your task is to build a queue using these two stacks.

🎯 Goal: Build a queue using two stacks in JavaScript. You will add items to the queue and remove them in the correct order (first in, first out) using only stack operations.

📋 What You'll Learn

Create two empty stacks named stackIn and stackOut

Create a variable queueSize to keep track of the number of items

Write a function enqueue that adds an item to stackIn and updates queueSize

Write a function dequeue that removes and returns the oldest item using stackOut and stackIn

Print the result of dequeuing after adding some items

💡 Why This Matters

🌍 Real World

Queues are used in many places like customer service lines, print job management, and task scheduling. Using two stacks to build a queue is a common programming exercise to understand data structures.

💼 Career

Understanding how to implement queues with stacks helps in coding interviews and improves problem-solving skills for software development roles.

1

Create two empty stacks

Create two empty arrays called stackIn and stackOut to represent the two stacks.

DSA Javascript

// Your code here

Hint

Use empty arrays to represent stacks in JavaScript.

2

Create a variable to track queue size

Create a variable called queueSize and set it to 0 to track the number of items in the queue.

DSA Javascript

const stackIn = [];
const stackOut = [];
// Your code here

Hint

Use let to create a variable that can change.

3

Write enqueue and dequeue functions

Write a function called enqueue that takes a parameter item, pushes it to stackIn, and increases queueSize by 1. Then write a function called dequeue that removes and returns the oldest item: if stackOut is empty, move all items from stackIn to stackOut by popping from stackIn and pushing to stackOut. Then pop from stackOut, decrease queueSize by 1, and return the popped item.

DSA Javascript

const stackIn = [];
const stackOut = [];
let queueSize = 0;
// Your code here

Hint

Remember to move items from stackIn to stackOut only when stackOut is empty.

4

Test the queue by enqueueing and dequeueing

Use enqueue to add 10, 20, and 30 to the queue. Then use dequeue to remove one item and print it using console.log.

DSA Javascript

const stackIn = [];
const stackOut = [];
let queueSize = 0;

function enqueue(item) {
    stackIn.push(item);
    queueSize++;
}

function dequeue() {
    if (stackOut.length === 0) {
        while (stackIn.length > 0) {
            stackOut.push(stackIn.pop());
        }
    }
    if (stackOut.length === 0) {
        return null; // Queue is empty
    }
    queueSize--;
    return stackOut.pop();
}

// Your code here

Hint

The first item enqueued should be the first item dequeued.

Practice

1. What is the main idea behind implementing a Queue using two stacks?

easy

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

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

C. Use two stacks to store elements in sorted order

D. Use one stack to hold even numbers and another for odd numbers

Solution

1. Step 1: Understand queue order

   A queue follows FIFO (First In First Out), so the first element added should be 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 A

4. Quick Check:

   Two stacks reverse order twice = Queue order [OK]

Hint: Two reversals with stacks keep queue order intact [OK]

Common Mistakes:

• Thinking one stack alone can maintain queue order
• Confusing stack sorting with queue behavior
• Using stacks for odd/even separation incorrectly

2. Which of the following is the correct syntax to push an element x onto a stack named stack1 in JavaScript?

easy

A. stack1.push(x);

B. stack1.add(x);

C. stack1.insert(x);

D. stack1.enqueue(x);

Solution

1. Step 1: Recall JavaScript stack push method

   JavaScript arrays use push() to add elements at the end, simulating stack push.

2. Step 2: Verify other options

   add(), insert(), and enqueue() are not valid array methods for pushing.

3. Final Answer:

   stack1.push(x); -> Option A

4. Quick Check:

   Array push method = stack1.push(x) [OK]

Hint: Use push() to add elements to stack in JavaScript [OK]

Common Mistakes:

• Using enqueue() which is for queues, not stacks
• Using add() or insert() which are not JavaScript array methods
• Confusing stack push with queue enqueue

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

class QueueTwoStacks {
  constructor() {
    this.stack1 = [];
    this.stack2 = [];
  }
  enqueue(x) {
    this.stack1.push(x);
  }
  dequeue() {
    if (this.stack2.length === 0) {
      while (this.stack1.length > 0) {
        this.stack2.push(this.stack1.pop());
      }
    }
    return this.stack2.pop();
  }
}

const q = new QueueTwoStacks();
q.enqueue(10);
q.enqueue(20);
q.enqueue(30);
console.log(q.dequeue());
q.enqueue(40);
console.log(q.dequeue());

medium

A. 20
30

B. 30
40

C. 10
30

D. 10
20

Solution

1. Step 1: Trace enqueue operations

   Elements 10, 20, 30 are pushed into stack1 in order: [10, 20, 30].

2. Step 2: Trace first dequeue

   stack2 is empty, so pop all from stack1 to stack2: stack2 becomes [30, 20, 10]. Then pop from stack2 returns 10.

3. Step 3: Enqueue 40 and second dequeue

   40 is pushed to stack1: [40]. stack2 still has [30, 20]. Dequeue pops from stack2 returning 20.

4. Final Answer:

   10 20 -> Option D

5. Quick Check:

   FIFO order maintained: 10 then 20 dequeued [OK]

Hint: Dequeue empties stack2 first, then pops top [OK]

Common Mistakes:

• Assuming dequeue pops directly from stack1
• Not transferring all elements to stack2 before dequeue
• Mixing order of elements after transfer

4. Identify the bug in this dequeue method of a queue implemented using two stacks:

dequeue() {
  while (this.stack1.length > 0) {
    this.stack2.push(this.stack1.pop());
  }
  return this.stack2.pop();
}

medium

A. It pops from stack1 instead of stack2

B. It never transfers elements from stack1 to stack2

C. It empties stack1 every dequeue without checking if stack2 is empty

D. It returns undefined always

Solution

1. Step 1: Analyze the transfer logic

   The method transfers all elements from stack1 to stack2 every time dequeue is called, without checking if stack2 is empty.

2. Step 2: Understand consequence

   This causes new elements to be appended to stack2 (which may have remaining older elements), violating FIFO as new elements are popped before older ones.

3. Final Answer:

   It empties stack1 every dequeue without checking if stack2 is empty -> Option C

4. Quick Check:

   Transfer only if stack2.length === 0 [OK]

Hint: Transfer only when stack2 is empty to keep elements safe [OK]

Common Mistakes:

• Transferring elements every dequeue call
• Not checking if stack2 is empty before transfer
• Confusing which stack to pop from

5. You want to implement a queue using two stacks where enqueue and dequeue operations are both efficient. Which approach below achieves amortized O(1) time per operation?

hard

A. Push elements alternately to stack1 and stack2 on enqueue; pop from stack1 on dequeue

B. Push all elements to stack1 on enqueue; transfer to stack2 only when stack2 is empty during dequeue

C. Push elements to stack1 and stack2 simultaneously on enqueue; pop from stack2 on dequeue

D. Push all elements to stack2 on enqueue; pop from stack1 on dequeue

Solution

1. Step 1: Understand amortized O(1) with two stacks

   Enqueue pushes to stack1 only. Dequeue pops from stack2. Transfer happens only when stack2 is empty, moving all elements from stack1 to stack2.

2. Step 2: Analyze other options

   Alternating pushes or pushing to both stacks simultaneously breaks queue order or wastes operations. Pushing all to stack2 and popping from stack1 is incorrect.

3. Final Answer:

   Push all elements to stack1 on enqueue; transfer to stack2 only when stack2 is empty during dequeue -> Option B

4. Quick Check:

   Lazy transfer on dequeue = amortized O(1) [OK]

Hint: Transfer only when needed during dequeue for efficiency [OK]

Common Mistakes:

• Transferring elements on every enqueue or dequeue
• Pushing to both stacks simultaneously
• Ignoring stack2 emptiness check before transfer