Queue Using Two Stacks in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Typescript

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

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

Apush

Bunshift

Cshift

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push to add elements.

Using shift or unshift which are array methods but not stack methods.

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

2fill in blank

medium

Complete the code to remove and return the front element from the queue.

DSA Typescript

dequeue(): number | undefined {
  if (this.stack2.length === 0) {
    while (this.stack1.length > 0) {
      this.stack2.[1](this.stack1.pop()!);
    }
  }
  return this.stack2.pop();
}

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

Ashift

Bpop

Cpush

Dunshift

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop on stack2 to transfer elements instead of push.

Using shift or unshift which are not stack methods.

We push elements popped from stack1 onto stack2 to reverse their order for dequeue.

3fill in blank

hard

Fix the error in the method that checks if the queue is empty.

DSA Typescript

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

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

A==

B>

C!=

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using != instead of == causing wrong empty check.

Using > or < which do not correctly check for zero length.

The queue is empty only if both stacks have zero length, so we check if stack1.length == 0.

4fill in blank

hard

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

DSA Typescript

peek(): number | undefined {
  if (this.stack2.length === 0) {
    while (this.stack1.length > 0) {
      this.stack2.[1](this.stack1.pop()!);
    }
  }
  return this.stack2[2];
}

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

Apush

Bpop()

C[this.stack2.length - 1]

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() to return the front element which removes it.

Accessing stack2[0] which is the bottom, not the top of the stack.

We push elements from stack1 to stack2, then return the last element in stack2 without popping it.

5fill in blank

hard

Fill all three blanks to implement the constructor initializing the two stacks as empty arrays.

DSA Typescript

constructor() {
  this.stack1 = [1];
  this.stack2 = [2];
  this.size = [3];
}

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

A[]

B0

Cnull

D{}

Attempts:

3 left

💡 Hint

Common Mistakes

Using null or {} instead of empty arrays for stacks.

Initializing size as null instead of zero.

Stacks are empty arrays and size starts at zero.

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