Queue Using Two Stacks in DSA Javascript - Execution Trace

Concept Flow - Queue Using Two Stacks

Enqueue Operation

↓

Push element onto stack1

↓

Done

↓

Dequeue Operation

↓

Is stack2 empty?

No→Pop element from stack2

Yes↓

Transfer all elements from stack1 to stack2

↓

Pop element from stack2

↓

Done

Enqueue adds elements to stack1. Dequeue removes from stack2. If stack2 is empty, transfer all from stack1 to stack2 first.

Execution Sample

DSA Javascript

enqueue(1)
enqueue(2)
dequeue()
enqueue(3)
dequeue()
dequeue()

This code adds 1 and 2, removes one element, adds 3, then removes two elements from the queue.

Execution Table

Step	Operation	stack1 (top→bottom)	stack2 (top→bottom)	Pointer Changes	Visual State (Queue front → rear)
1	enqueue(1)	[1]	[]	Push 1 to stack1	1 → null
2	enqueue(2)	[2, 1]	[]	Push 2 to stack1	1 → 2 → null
3	dequeue()	[]	[1, 2]	stack2 empty, transfer all from stack1 to stack2: stack1 emptied, stack2 = [1, 2]	1 → 2 → null
4	dequeue()	[]	[2]	Pop 1 from stack2	2 → null
5	enqueue(3)	[3]	[2]	Push 3 to stack1	2 → 3 → null
6	dequeue()	[3]	[]	Pop 2 from stack2	3 → null
7	dequeue()	[]	[3]	stack2 empty, transfer all from stack1 to stack2: stack1 emptied, stack2 = [3]	3 → null
8	dequeue()	[]	[]	Pop 3 from stack2	null

💡 All operations completed; queue is empty after last dequeue.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	After Step 8
stack1	[]	[1]	[2, 1]	[]	[]	[3]	[3]	[]	[]
stack2	[]	[]	[]	[1, 2]	[2]	[2]	[]	[3]	[]
Queue Size	0	1	2	2	1	2	1	1	0

Key Moments - 3 Insights

Why do we transfer all elements from stack1 to stack2 during dequeue?

Because stack2 holds elements in reverse order for dequeue. When stack2 is empty, transferring reverses stack1 order, so the oldest element is on top for correct queue behavior (see step 3 in execution_table).

Why don't we transfer elements on every dequeue?

We only transfer when stack2 is empty to avoid unnecessary moves. If stack2 has elements, we pop directly from it (see steps 4 and 6). This keeps dequeue efficient.

How does the queue front and rear relate to stacks?

The front of the queue is the top of stack2 if not empty; otherwise, elements are in stack1 waiting to be transferred. The rear is the top of stack1 (newest element). Visual states in execution_table show this.

Visual Quiz - 3 Questions

Look at the execution_table at step 3. What is the state of stack2 after transferring elements?

A [1, 2]

B [2, 1]

C []

D [3]

💡 Hint

Check the 'stack2 (top→bottom)' column at step 3 in execution_table.

At which step does the queue become empty?

A Step 6

B Step 8

C Step 7

D Step 5

💡 Hint

Look at the 'Visual State' column in execution_table to see when it shows 'null'.

If we enqueue(4) after step 8, what will be the state of stack1?

A []

B [3, 4]

C [4]

D [1, 4]

💡 Hint

Refer to variable_tracker for stack1 changes after enqueue operations.

Concept Snapshot

Queue Using Two Stacks:
- Enqueue: push element onto stack1
- Dequeue: if stack2 empty, transfer all from stack1 to stack2
- Then pop from stack2
- stack1 holds newest elements, stack2 holds oldest for dequeue
- Efficient queue using two LIFO stacks

Full Transcript

This visualization shows how a queue can be implemented using two stacks. Enqueue operations push elements onto stack1. Dequeue operations pop from stack2. If stack2 is empty during dequeue, all elements from stack1 are moved to stack2, reversing their order to maintain queue order. The execution table tracks each step, showing stack contents and queue state. Variable tracker shows how stack1 and stack2 change after each operation. Key moments clarify why transfer happens only when needed and how queue front and rear relate to stacks. The visual quiz tests understanding of stack states and queue emptiness. This method efficiently simulates queue behavior using two stacks.

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