Queue Using Two Stacks in DSA C# - 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.

How does the time to add or remove items grow as we add more items?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public class QueueUsingStacks {
    private Stack<int> stack1 = new Stack<int>();
    private Stack<int> stack2 = new Stack<int>();

    public void Enqueue(int x) {
        stack1.Push(x);
    }

    public int Dequeue() {
        if (stack2.Count == 0) {
            while (stack1.Count > 0) {
                stack2.Push(stack1.Pop());
            }
        }
        return stack2.Pop();
    }
}
    

This code adds items to one stack and removes items from another, moving items 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: Each item moves at most twice: once pushed to stack1, once popped and pushed to stack2.

How Execution Grows With Input

Adding items (enqueue) always takes the same small time per item.

Removing items (dequeue) sometimes takes longer when moving items between stacks, but this happens rarely.

Input Size (n)	Approx. Operations per Dequeue
10	About 1 or 10 operations occasionally
100	About 1 or 100 operations occasionally
1000	About 1 or 1000 operations occasionally

Pattern observation: Most dequeues are fast, but some take longer when stacks are moved. On average, time per operation stays steady.

Final Time Complexity

Time Complexity: O(1) amortized

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

Common Mistake

[X] Wrong: "Each dequeue always takes time proportional to the number of items because of moving stacks."

[OK] Correct: Items move only once from stack1 to stack2, so the cost spreads out over many dequeues, making average time constant.

Interview Connect

Understanding this helps you explain how clever use of data structures can keep operations fast on average, a skill valued in problem solving.

Self-Check

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

Practice

1. What is the main idea behind implementing a Queue using two stacks?

easy

A. Use one stack for enqueue and another for dequeue operations

B. Use two stacks to store elements in sorted order

C. Use two stacks to reverse the order twice to maintain queue order

D. Use two stacks to store only even and odd elements separately

Solution

1. Step 1: Understand queue order

   A queue follows First-In-First-Out (FIFO) order, meaning the first element added is 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, maintaining FIFO.

3. Final Answer:

   Use two stacks to reverse the order twice to maintain queue order -> Option C

4. Quick Check:

   Two reversals keep FIFO order [OK]

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

Common Mistakes:

• Thinking stacks store sorted elements
• Using stacks for separate data types
• Confusing enqueue and dequeue roles

2. Which of the following is the correct syntax to push an element 5 into a stack named stack1 in C#?

easy

A. stack1.Push(5);

B. stack1.Add(5);

C. stack1.Insert(5);

D. stack1.Enqueue(5);

Solution

1. Step 1: Recall stack push method

   In C#, the Stack<T> class uses the Push method to add elements.

2. Step 2: Check method names

   Add and Insert are not stack methods; Enqueue is for queues.

3. Final Answer:

   stack1.Push(5); -> Option A

4. Quick Check:

   Stack push method = Push() [OK]

Hint: Use Push() to add elements to stack [OK]

Common Mistakes:

• Using Add() or Insert() which are not stack methods
• Confusing Enqueue() with Push()
• Using wrong method names

3. What will be the output of the following C# code implementing a queue using two stacks?

var queue = new QueueUsingTwoStacks();
queue.Enqueue(10);
queue.Enqueue(20);
queue.Enqueue(30);
Console.WriteLine(queue.Dequeue());
queue.Enqueue(40);
Console.WriteLine(queue.Dequeue());

medium

A. 30\n40

B. 10\n20

C. 20\n30

D. 10\n30

Solution

1. Step 1: Enqueue elements 10, 20, 30

   Elements are pushed into the first stack in order: 10, then 20, then 30.

2. Step 2: Dequeue operation

   On dequeue, elements are popped from the second stack after transferring from the first, so 10 is dequeued first.

3. Step 3: Enqueue 40 and dequeue again

   40 is pushed into the first stack. Next dequeue returns 20, the next in queue order.

4. Final Answer:

   10 20 -> Option B

5. Quick Check:

   FIFO order maintained: 10 then 20 [OK]

Hint: Dequeue returns oldest enqueued element first [OK]

Common Mistakes:

• Assuming dequeue returns last pushed element
• Not transferring elements between stacks correctly
• Mixing enqueue and dequeue order

4. Identify the error in the following C# code snippet for dequeue operation in a queue using two stacks:

public int Dequeue()
{
    if (stack2.Count == 0)
    {
        while (stack1.Count > 0)
            stack2.Push(stack1.Pop());
    }
    return stack1.Pop();
}

medium

A. Should pop from stack2, not stack1, after transfer

B. Should check if stack1 is empty before popping

C. Should push elements back to stack1 after popping

D. Should use Peek() instead of Pop()

Solution

1. Step 1: Understand dequeue logic

   After transferring elements from stack1 to stack2, dequeue should pop from stack2 to maintain correct order.

2. Step 2: Identify wrong pop call

   The code incorrectly pops from stack1 instead of stack2 after transfer.

3. Final Answer:

   Should pop from stack2, not stack1, after transfer -> Option A

4. Quick Check:

   Pop from stack2 after transfer [OK]

Hint: Pop from second stack after transfer for dequeue [OK]

Common Mistakes:

• Popping from wrong stack after transfer
• Not checking if stacks are empty
• Using Peek() instead of Pop() for dequeue

5. You want to implement a queue using two stacks where Enqueue is O(1) time and Dequeue is amortized O(1). Which approach is best?

hard

A. Push elements to stack2 on enqueue; pop from stack1 on dequeue

B. Push elements alternately to stack1 and stack2 on enqueue; pop from stack1 on dequeue

C. Push elements to stack1 on enqueue; pop from stack1 on dequeue without transfer

D. Push all elements to stack1 on enqueue; transfer to stack2 only when stack2 is empty during dequeue

Solution

1. Step 1: Understand enqueue O(1) requirement

   Enqueue should just push to stack1 without extra operations for O(1) time.

2. Step 2: Understand dequeue amortized O(1)

   Dequeue transfers elements from stack1 to stack2 only when stack2 is empty, making transfers infrequent and amortized O(1).

3. Step 3: Evaluate options

   Push all elements to stack1 on enqueue; transfer to stack2 only when stack2 is empty during dequeue matches this approach; others do not maintain correct order or time complexity.

4. Final Answer:

   Push all elements to stack1 on enqueue; transfer to stack2 only when stack2 is empty during dequeue -> Option D

5. Quick Check:

   Lazy transfer on dequeue gives amortized O(1) [OK]

Hint: Transfer only when needed on dequeue for amortized O(1) [OK]

Common Mistakes:

• Transferring on every enqueue causing O(n) enqueue
• Popping from wrong stack without transfer
• Alternating pushes breaking queue order