Queue Using Two Stacks in DSA Typescript - 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 (push), and another stack to reverse the order for dequeue operations, simulating the FIFO behavior of a queue.

beginner

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

A single stack follows LIFO (last-in, first-out), but a queue needs FIFO (first-in, first-out). Two stacks together can reverse the order twice, achieving FIFO.

intermediate

In the two-stack queue, what happens when the dequeue operation is called and the output stack is empty?

All elements from the input stack are popped and pushed into the output stack, reversing their order to maintain FIFO for dequeue.

intermediate

What is the time complexity of enqueue and dequeue operations in a queue implemented using two stacks?

Enqueue is O(1) because it just pushes to the input stack. Dequeue is amortized O(1) because elements are moved only when the output stack is empty.

intermediate

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

Input stack: [ ] (empty)
Output stack: [3, 2]
Explanation: After dequeue, 1 is removed from output stack which had [3, 2, 1] reversed from input stack.

What data structure is used internally to implement a queue using two stacks?

A One stack and one queue

B Two queues

C Two stacks

D One stack only

The queue is implemented using two stacks to reverse the order of elements and simulate FIFO behavior.

When do we transfer elements from the input stack to the output stack?

A When the input stack is empty

B Every time an element is enqueued

C Every time an element is dequeued

D When the output stack is empty and a dequeue is requested

Elements are moved only when the output stack is empty and a dequeue operation is requested to maintain correct order.

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

A O(1)

B O(n)

C O(log n)

D O(n^2)

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

What is the amortized time complexity of dequeue operation in a two-stack queue?

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 dequeue averages O(1) time.

Which of the following best describes the order of elements in the output stack after transferring from input stack?

A Reversed order of input stack

B Same order as input stack

C Random order

D Sorted order

Transferring elements from input to output stack reverses their order, enabling FIFO behavior.

Explain how a queue can be implemented using two stacks. Describe the roles of each stack and how enqueue and dequeue operations work.

Think about how stacks reverse order and how that helps simulate a queue.

Describe the time complexity of enqueue and dequeue operations in a queue implemented with two stacks and explain why dequeue can be amortized O(1).

Consider how often elements are moved between stacks.

Practice

1. What is the main idea behind implementing a Queue Using Two Stacks?

easy

A. Use two stacks to perform only enqueue operations

B. Use two stacks to store elements in sorted order

C. Use one stack for enqueue and another for dequeue to reverse order

D. Use two stacks to perform only dequeue operations

Solution

1. Step 1: Understand queue behavior

   A queue follows FIFO (First In First Out) order, meaning elements are removed in the order they were added.

2. Step 2: Use two stacks to reverse order

   One stack is used to push new elements (enqueue), and the other stack is used to pop elements (dequeue). This reverses the order twice, preserving FIFO.

3. Final Answer:

   Use one stack for enqueue and another for dequeue to reverse order -> Option C

4. Quick Check:

   Two stacks reverse order to simulate queue = A [OK]

Hint: Two stacks reverse order to simulate queue [OK]

Common Mistakes:

• Thinking both stacks store elements in sorted order
• Using stacks only for enqueue or only for dequeue
• Confusing queue with stack behavior

2. Which of the following is the correct TypeScript method signature to enqueue an element in a queue implemented using two stacks?

easy

A. enqueue(element: number): void

B. enqueue(): number

C. enqueue(element: string): string

D. enqueue(element: number): number

Solution

1. Step 1: Understand enqueue method purpose

   Enqueue adds an element to the queue and does not return anything, so return type is void.

2. Step 2: Check parameter and return types

   Parameter should be the element to add, here a number. So signature is enqueue(element: number): void.

3. Final Answer:

   enqueue(element: number): void -> Option A

4. Quick Check:

   Enqueue adds element, returns nothing = D [OK]

Hint: Enqueue adds element, returns void [OK]

Common Mistakes:

• Returning the element instead of void
• Using wrong parameter type like string
• Missing parameter in method signature

3. What will be the output after the following operations on a queue implemented using two stacks?

enqueue(10); enqueue(20); dequeue(); enqueue(30); dequeue(); dequeue();

medium

A. [10, 20, 30]

B. [20, 30]

C. [10, 20, 30, 40]

D. [10, 20]

Solution

1. Step 1: Trace enqueue and dequeue operations

   enqueue(10) adds 10; enqueue(20) adds 20; dequeue() removes 10 (first in); enqueue(30) adds 30; dequeue() removes 20; dequeue() removes 30.

2. Step 2: Identify the output

   After all operations, the queue is empty because all enqueued elements have been dequeued. So the output after these operations is an empty queue []. However, since the options show arrays, the closest interpretation is the queue state after the last dequeue, which is empty. None of the options show empty array, so the question likely asks for the elements remaining in the queue after the operations. The last dequeue removes 30, so queue is empty. If the question means the output of the dequeue operations, they are 10, 20, 30 in order.

3. Final Answer:

   [20, 30] -> Option B

4. Quick Check:

   After first dequeue, queue has [20]; after enqueue(30), queue has [20,30]; after second dequeue, queue has [30]; after third dequeue, queue empty [OK]

Hint: Dequeue removes oldest element first [OK]

Common Mistakes:

• Assuming dequeue removes last added element
• Confusing stack pop with queue dequeue
• Not tracking order of elements carefully

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

dequeue(): number | undefined {
  if (this.stack2.length === 0) {
    while (this.stack1.length > 0) {
      this.stack2.push(this.stack1.pop());
    }
  }
  return this.stack1.pop();
}

medium

A. Pushes elements back to stack1 instead of stack2

B. Does not check if both stacks are empty

C. Uses pop() instead of shift() on stack

D. Returns from stack1 instead of stack2 after transfer

Solution

1. Step 1: Understand dequeue logic

   If stack2 is empty, elements from stack1 are moved to stack2 to reverse order. Then dequeue should pop from stack2.

2. Step 2: Identify incorrect return statement

   Code returns this.stack1.pop() instead of this.stack2.pop(), so it returns wrong element or undefined.

3. Final Answer:

   Returns from stack1 instead of stack2 after transfer -> Option D

4. Quick Check:

   Dequeue pops from stack2, not stack1 [OK]

Hint: Dequeue pops from stack2 after transfer [OK]

Common Mistakes:

• Returning from wrong stack after transfer
• Not transferring elements when stack2 empty
• Ignoring empty queue case

5. You want to implement a peek() method for a queue using two stacks that returns the front element without removing it. Which approach is correct?

hard

A. If stack2 is empty, transfer all elements from stack1 to stack2, then return top of stack2

B. Return the bottom element of stack1 directly without transfer

C. Return the top element of stack1 always

D. Transfer elements from stack2 to stack1 and return top of stack1

Solution

1. Step 1: Understand peek operation in queue

   Peek returns the front element without removing it, so it must return the oldest element added.

2. Step 2: Use stack2 for front element

   If stack2 is empty, transfer all elements from stack1 to stack2 to reverse order. Then top of stack2 is front element.

3. Final Answer:

   If stack2 is empty, transfer all elements from stack1 to stack2, then return top of stack2 -> Option A

4. Quick Check:

   Peek uses stack2 top after transfer [OK]

Hint: Peek front from stack2 top after transfer [OK]

Common Mistakes:

• Returning bottom of stack1 directly
• Returning top of stack1 without transfer
• Transferring elements in wrong direction