Queue Using Two Stacks in DSA Typescript - Execution Trace

Concept Flow - Queue Using Two Stacks

Enqueue Operation

↓

Push element onto Stack1

↓

Done

↓

Dequeue Operation

↓

Is Stack2 empty?

No→Pop from Stack2

Yes↓

Transfer all from Stack1 to Stack2

↓

Pop from Stack2

↓

Done

Enqueue pushes elements onto Stack1. Dequeue pops from Stack2; if Stack2 is empty, transfer all elements from Stack1 to Stack2 first.

Execution Sample

DSA Typescript

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

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

Execution Table

Step	Operation	Stack1 State	Stack2 State	Action	Visual State
1	enqueue(1)	[] -> [1]	[]	Push 1 to Stack1	Stack1: [1], Stack2: []
2	enqueue(2)	[1] -> [1,2]	[]	Push 2 to Stack1	Stack1: [1,2], Stack2: []
3	dequeue()	[1,2] -> []	[] -> [2,1]	Transfer all from Stack1 to Stack2 then pop from Stack2	Stack1: [], Stack2: [2,1]
4	enqueue(3)	[] -> [3]	[1]	Push 3 to Stack1	Stack1: [3], Stack2: [1]
5	dequeue()	[3]	[1] -> []	Pop from Stack2	Stack1: [3], Stack2: []
6	dequeue()	[3] -> []	[]	Transfer all from Stack1 to Stack2 then pop from Stack2	Stack1: [], Stack2: []

💡 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
Stack1	[]	[1]	[1,2]	[]	[3]	[3]	[]
Stack2	[]	[]	[]	[2,1]	[1]	[]	[]

Key Moments - 3 Insights

Why do we transfer all elements from Stack1 to Stack2 only when Stack2 is empty during dequeue?

Because Stack2 holds elements in reverse order for dequeue. Transferring only when Stack2 is empty preserves correct order and avoids unnecessary moves, as shown in steps 3 and 6 in the execution_table.

What happens if we try to dequeue when both stacks are empty?

The queue is empty, so dequeue cannot remove any element. This situation is not shown in the current execution_table but would stop operation or return an error.

Why does enqueue always push to Stack1 and never to Stack2?

Enqueue pushes to Stack1 to keep new elements in order. Stack2 is only for dequeue operations, as seen in steps 1, 2, and 4 where enqueue pushes to Stack1.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the state of Stack2 after the operation?

A [2,1]

B [2]

C []

D [1]

💡 Hint

Refer to the 'Stack2 State' column in row for step 3 in execution_table.

At which step does Stack1 become empty for the first time?

A Step 2

B Step 4

C Step 3

D Step 5

💡 Hint

Check 'Stack1 State' column in execution_table rows to see when it changes to [].

If we enqueue(4) after step 6, what will be the state of Stack1 and Stack2?

A Stack1: [], Stack2: [4]

B Stack1: [4], Stack2: []

C Stack1: [3,4], Stack2: []

D Stack1: [], Stack2: []

💡 Hint

Enqueue always pushes to Stack1 as shown in steps 1, 2, and 4.

Concept Snapshot

Queue Using Two Stacks:
- Enqueue: push element onto Stack1
- Dequeue: if Stack2 empty, transfer all from Stack1 to Stack2, then pop Stack2
- Stack1 stores new elements; Stack2 reverses order for dequeue
- Efficient amortized queue using two stacks
- Maintains FIFO order using 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 elements 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 traces each step, showing stack states and actions. Variable tracker follows Stack1 and Stack2 contents after each operation. Key moments clarify why transfer happens only when Stack2 is empty, what happens on empty dequeue, and why enqueue always uses Stack1. The visual quiz tests understanding of stack states and operation effects. This method efficiently simulates a queue using two last-in-first-out 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