Queue Using Two Stacks in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to add an element to the queue using stackPush.

DSA Java

public void enqueue(int x) {
    stackPush.[1](x);
}

Drag options to blanks, or click blank then click option'

Aempty

Bpush

Cpeek

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push to add elements.

Trying to peek or check empty instead of adding.

To add an element to the queue, we push it onto stackPush using the push method.

2fill in blank

medium

Complete the code to move elements from stackPush to stackPop when stackPop is empty.

DSA Java

if (stackPop.[1]()) {
    while (!stackPush.empty()) {
        stackPop.push(stackPush.pop());
    }
}

Drag options to blanks, or click blank then click option'

Apop

Bpush

Cempty

Dpeek

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() to check emptiness instead of empty().

Using peek() which only looks at the top element.

We check if stackPop is empty using the empty() method before moving elements.

3fill in blank

hard

Fix the error in the dequeue method to return the correct element.

DSA Java

public int dequeue() {
    if (stackPop.empty()) {
        while (!stackPush.empty()) {
            stackPop.push(stackPush.pop());
        }
    }
    return stackPop.[1]();
}

Drag options to blanks, or click blank then click option'

Apop

Bpush

Cpeek

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek() which does not remove the element.

Using push() or empty() which are incorrect here.

To remove and return the front element of the queue, we pop from stackPop.

4fill in blank

hard

Fill both blanks to implement the peek method that returns the front element without removing it.

DSA Java

public int peek() {
    if (stackPop.[1]()) {
        while (!stackPush.empty()) {
            stackPop.push(stackPush.pop());
        }
    }
    return stackPop.[2]();
}

Drag options to blanks, or click blank then click option'

Aempty

Bpop

Cpeek

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() which removes the element instead of peek().

Not checking if stackPop is empty before moving elements.

We check if stackPop is empty using empty(), then return the top element using peek() without removing it.

5fill in blank

hard

Fill all three blanks to implement the isEmpty method that checks if the queue is empty.

DSA Java

public boolean isEmpty() {
    return stackPush.[1]() && stackPop.[2]();
}

// Usage example:
// Queue is empty if both stacks are empty: stackPush.[3]() and stackPop.empty()

Drag options to blanks, or click blank then click option'

Aempty

Bpop

Cpeek

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() or peek() instead of empty() to check emptiness.

Checking only one stack instead of both.

The queue is empty only if both stackPush and stackPop are empty, checked by empty() method.

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