Queue Using Two Stacks in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to add an element to the queue using stack1.

DSA Javascript

enqueue(element) {
  this.stack1.[1](element);
}

Drag options to blanks, or click blank then click option'

Aunshift

Bpush

Cshift

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'pop' which removes an element instead of adding.

Using 'shift' or 'unshift' which are for queue operations, not stack.

To add an element to the queue, we push it onto stack1.

2fill in blank

medium

Complete the code to check if the queue is empty.

DSA Javascript

isEmpty() {
  return this.stack1.length === [1] && this.stack2.length === 0;
}

Drag options to blanks, or click blank then click option'

A1

B-1

C0

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for length equal to 1 instead of 0.

Using null instead of zero for length comparison.

The queue is empty if both stacks have zero length.

3fill in blank

hard

Fix the error in the dequeue method to correctly move elements from stack1 to stack2.

DSA Javascript

dequeue() {
  if (this.stack2.length === 0) {
    while (this.stack1.length > [1]) {
      this.stack2.push(this.stack1.pop());
    }
  }
  return this.stack2.pop();
}

Drag options to blanks, or click blank then click option'

A0

Bnull

C-1

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using length > 1 which skips the last element.

Using negative or null values in the condition.

We move elements while stack1 is not empty, so length > 0.

4fill in blank

hard

Fill both blanks to complete the peek method that returns the front element without removing it.

DSA Javascript

peek() {
  if (this.stack2.length === 0) {
    while (this.stack1.length > [1]) {
      this.stack2.[2](this.stack1.pop());
    }
  }
  return this.stack2[this.stack2.length - 1];
}

Drag options to blanks, or click blank then click option'

A0

B1

Cpush

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'pop' instead of 'push' when moving elements.

Using length > 1 which misses elements.

We move elements while stack1 is not empty (length > 0) and push them onto stack2.

5fill in blank

hard

Fill all three blanks to complete the size method that returns the total number of elements in the queue.

DSA Javascript

size() {
  return this.stack1.[1] + this.stack2.[2];
}

// Example usage:
const q = new Queue();
q.enqueue(1);
q.enqueue(2);
q.dequeue();
console.log(q.size()); // Should print [3]

Drag options to blanks, or click blank then click option'

Alength

Bsize

C1

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'size' which is not a JavaScript array property.

Incorrect expected output number.

The size is the sum of lengths of both stacks. After enqueue 1, 2 and dequeue once, stack1 is empty (length 0) and stack2 has one element (length 1), total size is 1.

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