Queue Using Two Stacks in DSA Typescript - Time & Space Complexity

Time Complexity: Queue Using Two Stacks

O(1) amortized

Understanding Time Complexity

We want to understand how fast operations run when we build a queue using two stacks.

Specifically, how does the time to add or remove items grow as the queue gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class QueueUsingStacks {
  private stack1: number[] = [];
  private stack2: number[] = [];

  enqueue(value: number): void {
    this.stack1.push(value);
  }

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

This code adds items to one stack and removes items from another, moving items between stacks only when needed.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Moving all items from stack1 to stack2 when stack2 is empty during dequeue.
• How many times: This move happens only when stack2 is empty, not on every dequeue.

How Execution Grows With Input

When adding items, each enqueue is quick and simple.

When removing items, sometimes we move many items at once, but not every time.

Input Size (n)	Approx. Operations
10	About 10 pushes for enqueue, and up to 10 moves for dequeue when stack2 is empty.
100	About 100 pushes for enqueue, and up to 100 moves for dequeue occasionally.
1000	About 1000 pushes for enqueue, and up to 1000 moves for dequeue occasionally.

Pattern observation: Most operations are quick, but occasionally a bigger move happens, spreading the cost over many operations.

Final Time Complexity

Time Complexity: O(1) amortized

This means each operation takes constant time on average, even if some individual operations take longer.

Common Mistake

[X] Wrong: "Each dequeue always takes O(n) time because of moving items between stacks."

[OK] Correct: The moving happens only when stack2 is empty, so the cost spreads out over many dequeues, making average time per operation constant.

Interview Connect

Understanding this helps you explain how clever data structure design can make slow steps rare and keep average speed fast.

Self-Check

What if we moved items from stack1 to stack2 on every dequeue, even if stack2 is not empty? How would the time complexity change?

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