Queue Using Two Stacks in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of enqueue and dequeue operations

What is the output of the following Java code that uses two stacks to implement a queue?

DSA Java

import java.util.Stack;

class QueueUsingStacks {
    Stack<Integer> stack1 = new Stack<>();
    Stack<Integer> stack2 = new Stack<>();

    void enqueue(int x) {
        stack1.push(x);
    }

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

    public static void main(String[] args) {
        QueueUsingStacks queue = new QueueUsingStacks();
        queue.enqueue(10);
        queue.enqueue(20);
        int first = queue.dequeue();
        queue.enqueue(30);
        int second = queue.dequeue();
        int third = queue.dequeue();
        System.out.println(first + "," + second + "," + third);
    }
}

A 20,10,30

B 10,30,20

C 10,20,30

D 30,20,10

💡 Hint

Remember that dequeue transfers elements from stack1 to stack2 only when stack2 is empty.

The queue inserts 10 and 20 into stack1. On first dequeue, stack2 is empty, so elements move from stack1 to stack2 in reverse order: 20 then 10. The top of stack2 is 10, so 10 is dequeued first.

Then 30 is enqueued into stack1.

Second dequeue pops 20 from stack2.

Third dequeue finds stack2 empty, moves 30 from stack1 to stack2, then pops 30.

🧠 Conceptual

intermediate

Understanding time complexity of dequeue operation

What is the average time complexity of the dequeue operation in a queue implemented using two stacks?

A O(1) amortized time

B O(n) for every dequeue

C O(log n) for every dequeue

D O(n^2) in worst case

💡 Hint

Consider how often elements are moved between stacks.

Each element is pushed and popped at most twice: once when enqueued and once when moved between stacks. This means the average cost per dequeue is constant time, O(1) amortized.

🔧 Debug

advanced

Identify the error in dequeue implementation

What error will occur when running this dequeue method in Java if both stacks are empty?

DSA Java

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

A NullPointerException

B EmptyStackException

C IndexOutOfBoundsException

D No error, returns null

💡 Hint

What happens if you pop from an empty stack in Java?

If both stacks are empty, stack2.pop() tries to pop from an empty stack, causing an EmptyStackException.

❓ Predict Output

advanced

Output after mixed operations

What is the output of the following Java code snippet?

DSA Java

import java.util.Stack;

class QueueUsingStacks {
    Stack<Integer> s1 = new Stack<>();
    Stack<Integer> s2 = new Stack<>();

    void enqueue(int x) {
        s1.push(x);
    }

    int dequeue() {
        if (s2.isEmpty()) {
            while (!s1.isEmpty()) {
                s2.push(s1.pop());
            }
        }
        return s2.isEmpty() ? -1 : s2.pop();
    }

    public static void main(String[] args) {
        QueueUsingStacks q = new QueueUsingStacks();
        q.enqueue(5);
        q.enqueue(15);
        System.out.println(q.dequeue());
        q.enqueue(25);
        System.out.println(q.dequeue());
        System.out.println(q.dequeue());
        System.out.println(q.dequeue());
    }
}

A

5
15
-1
25

B

5
25
15
-1

C

15
5
25
-1

D

5
15
25
-1

💡 Hint

Check the order elements are dequeued after transfers between stacks.

First dequeue returns 5 (front of queue).

Then 25 is enqueued.

Second dequeue returns 15.

Third dequeue returns 25.

Fourth dequeue returns -1 because queue is empty.

🧠 Conceptual

expert

Space complexity of queue using two stacks

What is the worst-case space complexity of a queue implemented using two stacks when it contains n elements?

A O(n)

B O(n log n)

C O(2n)

D O(1)

💡 Hint

Consider how many elements can be stored in both stacks combined.

The n elements are distributed between the two stacks, so total space complexity is O(n).

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