Queue Using Two Stacks in DSA Java - Build from Scratch

Queue Using Two Stacks

📖 Scenario: Imagine you are managing a line of customers waiting for service. You want to keep track of who comes first and who goes next. A queue is perfect for this because it works like a real line: first in, first out.But you only have two stacks, which work like piles of plates (last in, first out). Your task is to use these two stacks to build a queue that behaves correctly.

🎯 Goal: Build a queue using two stacks in Java. You will add items to the queue and remove them in the correct order, using only stack operations.

📋 What You'll Learn

Create two stacks using Java's Stack class

Implement enqueue method to add elements to the queue

Implement dequeue method to remove elements from the queue

Print the queue elements in order after operations

💡 Why This Matters

🌍 Real World

Queues are used in many places like customer service lines, task scheduling, and buffering data streams. Using stacks to build queues helps understand how data structures can work together.

💼 Career

Understanding how to implement one data structure using others is a common interview question and shows problem-solving skills important for software development roles.

1

Create two stacks for the queue

Create two stacks called stack1 and stack2 using Stack<Integer>.

DSA Java

// Your code here

Hint

Use Stack<Integer> to create two empty stacks named stack1 and stack2.

2

Add an enqueue method

Add a method called enqueue that takes an int parameter called value and pushes it onto stack1.

DSA Java

import java.util.Stack;

public class QueueUsingTwoStacks {
    Stack stack1 = new Stack();
    Stack stack2 = new Stack();

    // Your code here
}

Hint

Use stack1.push(value) inside the enqueue method to add the value.

3

Add a dequeue method

Add a method called dequeue that returns an int. If stack2 is empty, pop all elements from stack1 and push them onto stack2. Then pop and return the top element from stack2. If both stacks are empty, return -1.

DSA Java

import java.util.Stack;

public class QueueUsingTwoStacks {
    Stack stack1 = new Stack();
    Stack stack2 = new Stack();

    public void enqueue(int value) {
        stack1.push(value);
    }

    // Your code here
}

Hint

Move elements from stack1 to stack2 only if stack2 is empty, then pop from stack2.

4

Print queue elements after operations

In the main method, create an instance of QueueUsingTwoStacks. Enqueue 10, 20, and 30. Dequeue one element. Then enqueue 40. Finally, dequeue all elements one by one and print each dequeued element on its own line.

DSA Java

import java.util.Stack;

public class QueueUsingTwoStacks {
    Stack stack1 = new Stack();
    Stack stack2 = new Stack();

    public void enqueue(int value) {
        stack1.push(value);
    }

    public int dequeue() {
        if (stack2.isEmpty()) {
            while (!stack1.isEmpty()) {
                stack2.push(stack1.pop());
            }
        }
        if (stack2.isEmpty()) {
            return -1;
        }
        return stack2.pop();
    }

    public static void main(String[] args) {
        // Your code here
    }
}

Hint

Follow the order: enqueue 10, 20, 30; dequeue once; enqueue 40; then dequeue all and print each.

Practice

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

easy

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

B. Use two stacks to store elements in sorted order

C. Use one stack and a linked list to simulate queue behavior

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

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

   Stacks are Last-In-First-Out (LIFO). Using two stacks, we reverse the order twice, restoring FIFO behavior.

3. Final Answer:

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

4. Quick Check:

   Two stacks reverse twice = queue order [OK]

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

Common Mistakes:

• Thinking one stack alone can maintain queue order
• Assuming stacks keep elements sorted
• Confusing enqueue and dequeue operations

2. Which of the following is the correct Java method signature for enqueue operation in a queue implemented using two stacks?

easy

A. public int enqueue(int x)

B. void enqueue(int x)

C. public void enqueue() int x

D. public void enqueue(int x)

Solution

1. Step 1: Identify correct Java method signature

   In Java, method signature includes access modifier, return type, method name, and parameters. Enqueue adds an element and returns nothing, so return type is void.

2. Step 2: Check each option

   public void enqueue(int x) has public access, void return type, method name enqueue, and int parameter. This is correct syntax.

3. Final Answer:

   public void enqueue(int x) -> Option D

4. Quick Check:

   Method signature with void return and int parameter [OK]

Hint: Enqueue returns void and takes element parameter [OK]

Common Mistakes:

• Missing return type or wrong return type
• Incorrect parameter placement
• Syntax errors like misplaced braces

3. Consider the following Java code snippet implementing dequeue operation using two stacks. What will be the output after these operations?

QueueUsingTwoStacks q = new QueueUsingTwoStacks();
q.enqueue(10);
q.enqueue(20);
q.enqueue(30);
System.out.println(q.dequeue());
q.enqueue(40);
System.out.println(q.dequeue());

medium

A. 10\n20

B. 30\n10

C. 10\n30

D. 20\n30

Solution

1. Step 1: Trace enqueue operations

   Elements enqueued in order: 10, 20, 30. Queue order is [10, 20, 30].

2. Step 2: Dequeue and enqueue operations

   First dequeue removes 10 (front). Then enqueue 40, queue is [20, 30, 40]. Second dequeue removes 20.

3. Final Answer:

   10 20 -> Option A

4. Quick Check:

   Dequeue order matches enqueue FIFO [OK]

Hint: Dequeue returns oldest enqueued element [OK]

Common Mistakes:

• Assuming stack order instead of queue order
• Mixing up enqueue and dequeue sequence
• Forgetting to enqueue after dequeue

4. The following code snippet for dequeue operation in a queue using two stacks throws an error. What is the error and how to fix it?

public int dequeue() {
  if (stack2.isEmpty()) {
    while (!stack1.isEmpty()) {
      stack2.push(stack1.pop());
    }
  }
  return stack1.pop();
}

medium

A. Missing check for empty queue before popping

B. Returns from wrong stack; should pop from stack2

C. Should push from stack2 to stack1 instead

D. Syntax error: missing semicolon after while loop

Solution

1. Step 1: Analyze dequeue logic

   When stack2 is empty, elements are moved from stack1 to stack2 to reverse order. Dequeue should pop from stack2 to get the front element.

2. Step 2: Identify error in return statement

   Code returns stack1.pop(), which is incorrect. It should return stack2.pop() to dequeue correctly.

3. Final Answer:

   Returns from wrong stack; should pop from stack2 -> Option B

4. Quick Check:

   Dequeue pops from stack2 after transfer [OK]

Hint: Dequeue pops from stack2 after transfer [OK]

Common Mistakes:

• Popping from stack1 instead of stack2
• Not transferring elements when stack2 is empty
• Ignoring empty queue condition

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?

Options:
A) Push new elements to stack1 on enqueue; on dequeue, if stack2 empty, move all from stack1 to stack2, then pop stack2.
B) Push new elements to stack2 on enqueue; on dequeue, pop from stack1.
C) Push new elements to stack1 on enqueue; on dequeue, pop from stack1 directly.
D) Push new elements to stack2 on enqueue; on dequeue, if stack1 empty, move all from stack2 to stack1, then pop stack1.

hard

A. Push to stack1 on enqueue; dequeue from stack1 directly

B. Push to stack2 on enqueue; dequeue from stack1 directly

C. Push to stack1 on enqueue; transfer to stack2 on dequeue if empty

D. Push to stack2 on enqueue; transfer to stack1 on dequeue if empty

Solution

1. Step 1: Understand enqueue and dequeue costs

   Enqueue should be O(1), so pushing to a stack directly is best. Dequeue can be amortized O(1) by transferring elements only when needed.

2. Step 2: Identify correct approach

   Push to stack1 on enqueue; transfer to stack2 on dequeue if empty pushes to stack1 on enqueue (O(1)). On dequeue, if stack2 is empty, transfer all from stack1 to stack2 (reversing order), then pop from stack2. This gives amortized O(1) dequeue.

3. Final Answer:

   Push to stack1 on enqueue; transfer to stack2 on dequeue if empty -> Option C

4. Quick Check:

   Enqueue O(1), dequeue amortized O(1) with two stacks [OK]

Hint: Enqueue to stack1; dequeue from stack2 after transfer [OK]

Common Mistakes:

• Pushing to wrong stack on enqueue
• Popping from wrong stack on dequeue
• Not transferring elements when needed