Queue Using Two Stacks in DSA Javascript - 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.

intermediate

How does the dequeue operation work in a queue implemented with two stacks?

If the second stack is empty, pop all elements from the first stack and push them into the second stack. Then pop from the second stack to get the front element of the queue.

beginner

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

One stack alone cannot maintain the FIFO order because it is LIFO. Using two stacks allows reversing the order twice, which restores the FIFO order needed for a queue.

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 first stack. Dequeue is amortized O(1) because elements are moved between stacks only when needed.

intermediate

Show the state of two stacks after enqueueing 1, 2, 3 and then dequeueing one element.

After enqueue 1,2,3: stack1 = [1,2,3], stack2 = []. After dequeue: stack1 = [], stack2 = [3,2], dequeued element = 1.

Which data structure is used to reverse the order of elements in a queue implemented with two stacks?

A Queue

B Stack

C Linked List

D Array

Stacks reverse the order of elements when popping and pushing, which helps simulate queue behavior.

What happens when the dequeue stack is empty during a dequeue operation?

A Elements are popped from enqueue stack and pushed into dequeue stack

B Queue is empty

C Elements are pushed into enqueue stack

D Nothing happens

To get the front element, elements are moved from the enqueue stack to the dequeue stack to reverse their order.

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

A O(n^2)

B O(1)

C O(log n)

D O(n)

In the worst case, all elements are moved from one stack to another, which takes O(n) time.

What is the amortized time complexity of dequeue operation in a queue using two stacks?

A O(1)

B O(n)

C O(log n)

D O(n^2)

Amortized over many operations, each element is moved at most once, so average dequeue is O(1).

Which operation is always O(1) in a queue implemented with two stacks?

A Dequeue

B Both Enqueue and Dequeue

C Enqueue

D Neither

Enqueue just pushes an element onto the first stack, which is always O(1).

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

Think about reversing order with stacks to get FIFO.

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

Consider when elements move between 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