Implement Stack Using Queue in DSA C# - Time & Space Complexity

Time Complexity: Implement Stack Using Queue

O(n)

Understanding Time Complexity

When we build a stack using a queue, we want to know how fast the main actions like push and pop happen.

We ask: How does the time to do these actions grow as we add more items?

Scenario Under Consideration

Analyze the time complexity of the push operation in this stack using one queue.

public class StackUsingQueue {
    private Queue<int> queue = new Queue<int>();

    public void Push(int x) {
        queue.Enqueue(x);
        int size = queue.Count;
        while (size > 1) {
            queue.Enqueue(queue.Dequeue());
            size--;
        }
    }
}
    

This code adds an item to the stack by putting it in the queue and then rotating the queue to keep the newest item at the front.

Identify Repeating Operations

Look at what repeats when we push:

• Primary operation: Moving items inside the queue by dequeueing and enqueueing.
• How many times: This happens once for each item already in the queue, so it repeats n-1 times if there are n items.

How Execution Grows With Input

As we add more items, the push operation takes longer because it moves all existing items each time.

Input Size (n)	Approx. Operations
10	About 9 moves inside the queue
100	About 99 moves inside the queue
1000	About 999 moves inside the queue

Pattern observation: The work grows directly with the number of items, so doubling items doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means pushing an item takes time proportional to how many items are already in the stack.

Common Mistake

[X] Wrong: "Push in this stack is always fast because queue operations are fast."

[OK] Correct: Even though queue operations are quick, we do many of them during push, so the total time grows with the number of items.

Interview Connect

Understanding how to build one data structure using another and analyzing its speed shows your problem-solving skills and helps you think about trade-offs clearly.

Self-Check

"What if we used two queues instead of one? How would the time complexity of push and pop change?"

Practice

1. What is the main idea behind implementing a stack using a queue?

easy

A. Use a linked list directly to implement stack

B. Use one queue and reorder elements to simulate stack behavior

C. Use two stacks to simulate queue behavior

D. Use an array to simulate queue behavior

Solution

1. Step 1: Understand stack and queue behavior

   A stack follows Last-In-First-Out (LIFO), while a queue follows First-In-First-Out (FIFO).

2. Step 2: Simulate stack using queue

   By reordering elements in a single queue after each push, the newest element is always at the front, simulating stack pop behavior.

3. Final Answer:

   Use one queue and reorder elements to simulate stack behavior -> Option B

4. Quick Check:

   Stack via queue = reorder elements [OK]

Hint: Remember stack is LIFO, queue is FIFO; reorder queue to mimic stack [OK]

Common Mistakes:

• Confusing stack and queue order
• Using two stacks instead of queues
• Ignoring element reordering after push

2. Which of the following is the correct C# syntax to declare a queue of integers?

easy

A. Queue<int> queue = new Queue<int>();

B. Queue queue = new Queue<int>();

C. Queue<int> queue = Queue<int>();

D. Queue<int> queue = new Queue();

Solution

1. Step 1: Review C# generic queue declaration

   In C#, a generic queue is declared as Queue<T> with new keyword and parentheses.

2. Step 2: Check each option

   Queue<int> queue = new Queue<int>(); correctly uses Queue<int> queue = new Queue<int>(); which is valid syntax.

3. Final Answer:

   Queue<int> queue = new Queue<int>(); -> Option A

4. Quick Check:

   Generic queue syntax = Queue<int> queue = new Queue<int>(); [OK]

Hint: Use new Queue<int>() to declare generic queue in C# [OK]

Common Mistakes:

• Omitting new keyword
• Missing generic type in constructor
• Using non-generic Queue without type

3. What will be the output of the following C# code implementing stack using one queue?

var stack = new StackUsingQueue();
stack.Push(10);
stack.Push(20);
stack.Push(30);
Console.WriteLine(stack.Pop());
Console.WriteLine(stack.Pop());

medium

A. 10\n30

B. 10\n20

C. 20\n30

D. 30\n20

Solution

1. Step 1: Understand push operation in stack using queue

   Each push adds element and reorders queue so last pushed is at front.

2. Step 2: Trace pop operations

   First pop returns 30 (last pushed), second pop returns 20 (second last pushed).

3. Final Answer:

   30\n20 -> Option D

4. Quick Check:

   Stack pop order = last in first out = 30, then 20 [OK]

Hint: Pop returns last pushed element due to queue reorder [OK]

Common Mistakes:

• Assuming queue FIFO order for pop
• Mixing push and pop order
• Ignoring reorder after push

4. Identify the potential issue in this C# method for stack push using one queue:

public void Push(int x) {
    queue.Enqueue(x);
    for (int i = 0; i < queue.Count - 1; i++) {
        queue.Enqueue(queue.Dequeue());
    }
}

medium

A. Enqueue should be Dequeue

B. Missing return statement

C. Using queue.Count in loop condition changes during iteration

D. Loop should run queue.Count times

Solution

1. Step 1: Analyze loop condition with queue.Count

   queue.Count changes momentarily during iteration (decreases after Dequeue, restores after Enqueue).

2. Step 2: Explain why this is a concern

   Although the size restores before the next condition check and works correctly, using a changing value in the loop condition can be confusing and is bad practice. Store count beforehand: int n = queue.Count; for(int i=0; i < n-1; i++)...

3. Final Answer:

   Using queue.Count in loop condition changes during iteration -> Option C

4. Quick Check:

   Loop count must be stored before loop [OK]

Hint: Store queue.Count before loop to avoid changing condition [OK]

Common Mistakes:

• Using dynamic queue.Count in loop condition
• Not rotating queue elements correctly
• Confusing enqueue and dequeue operations

5. You want to implement a stack using two queues. Which approach correctly simulates the stack's pop operation?

hard

A. Push elements into queue1; for pop, move all but last from queue1 to queue2, dequeue last from queue1, then swap queues

B. Push elements into queue2; for pop, dequeue directly from queue2

C. Push elements into queue1; pop by dequeuing from queue1 without moving elements

D. Push elements into queue2; for pop, move all elements to queue1 and dequeue first

Solution

1. Step 1: Understand two-queue stack implementation

   One queue holds elements; pop requires removing last pushed element, which is at the end.

2. Step 2: Describe correct pop operation

   Move all but last element from queue1 to queue2, dequeue last from queue1 (stack pop), then swap queues for next operations.

3. Final Answer:

   Push elements into queue1; for pop, move all but last from queue1 to queue2, dequeue last from queue1, then swap queues -> Option A

4. Quick Check:

   Pop = move all but last, dequeue last, swap queues [OK]

Hint: Pop by moving all but last element, then dequeue last [OK]

Common Mistakes:

• Dequeuing directly without moving elements
• Not swapping queues after pop
• Confusing which queue holds elements