Implement Stack Using Queue in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to add an element to the stack using a queue.

DSA Javascript

push(x) {
  this.queue.push(x);
  let size = this.queue.length;
  while (size > 1) {
    this.queue.push(this.queue.[1]());
    size--;
  }
}

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

Apop

Bslice

Cunshift

Dshift

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() removes the last element, which breaks the queue order.

Using unshift() or slice() does not remove elements.

The shift() method removes the first element from the queue, which is then pushed back to the end to maintain stack order.

2fill in blank

medium

Complete the code to remove the top element from the stack.

DSA Javascript

pop() {
  if (this.queue.length === 0) return null;
  return this.queue.[1]();
}

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

Aslice

Bshift

Cpop

Dunshift

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() removes the last element, not the top of the stack here.

Returning null without checking queue length causes errors.

The shift() method removes and returns the first element, which is the top of the stack in this implementation.

3fill in blank

hard

Fix the error in the peek method to return the top element without removing it.

DSA Javascript

peek() {
  if (this.queue.length === 0) return null;
  return this.queue[[1]];
}

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

A-1

Bthis.queue.length - 1

C0

Dthis.queue.length

Attempts:

3 left

💡 Hint

Common Mistakes

Using last index returns the bottom element, not the top.

Using negative or out-of-range indices causes errors.

The top element is at index 0 because the queue is rotated after each push to keep the top at the front.

4fill in blank

hard

Fill both blanks to check if the stack is empty.

DSA Javascript

isEmpty() {
  return this.queue.[1] === [2];
}

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

Alength

Bsize

C0

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'size' instead of 'length' property.

Comparing length to null instead of zero.

The stack is empty if the queue's length is equal to 0.

5fill in blank

hard

Fill all three blanks to implement the stack class constructor and initialize the queue.

DSA Javascript

class Stack {
  constructor() {
    this.queue = [1];
  }

  push(x) {
    this.queue.push(x);
    let size = this.queue.length;
    while (size > 1) {
      this.queue.push(this.queue.[2]());
      size--;
    }
  }

  pop() {
    if (this.queue.length === 0) return null;
    return this.queue.[3]();
  }
}

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

A[]

Bshift

Cpop

D{}

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing queue as an object {} instead of array [].

Using pop() in push rotation breaks stack order.

Using shift() in pop method removes wrong element.

The queue is an array initialized with []. The shift() method removes the first element for push rotation, and pop() is used in pop to remove the top element.

Practice

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

easy

A. Use two stacks to implement a queue

B. Use an array to implement a queue

C. Use a linked list to implement a stack

D. Use a queue to simulate stack operations by reversing order

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

   To implement stack using queue, we reverse the order of elements in the queue to mimic LIFO behavior.

3. Final Answer:

   Use a queue to simulate stack operations by reversing order -> Option D

4. Quick Check:

   Stack via queue = reverse order [OK]

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

Common Mistakes:

• Confusing stack with queue behavior
• Trying to implement queue using stack instead
• Ignoring order reversal needed for stack simulation

2. Which of the following is the correct syntax to enqueue an element `x` into a JavaScript queue implemented as an array named `queue`?

easy

A. queue.shift(x);

B. queue.push(x);

C. queue.pop(x);

D. queue.unshift(x);

Solution

1. Step 1: Identify enqueue operation in JavaScript array

   Enqueue means adding element at the end of the queue, which is done by push().

2. Step 2: Check other methods

   pop() removes from end, shift() removes from front, unshift() adds at front, so only push() adds at end.

3. Final Answer:

   queue.push(x); -> Option B

4. Quick Check:

   Enqueue = push() [OK]

Hint: Use push() to add at queue end in JavaScript arrays [OK]

Common Mistakes:

• Using pop() instead of push() to add elements
• Using shift() or unshift() incorrectly for enqueue
• Confusing enqueue with dequeue operations

3. Consider this JavaScript code implementing a stack using one queue:

class Stack {
  constructor() {
    this.queue = [];
  }
  push(x) {
    this.queue.push(x);
    for (let i = 0; i < this.queue.length - 1; i++) {
      this.queue.push(this.queue.shift());
    }
  }
  pop() {
    return this.queue.shift();
  }
}
const s = new Stack();
s.push(1);
s.push(2);
s.push(3);
console.log(s.pop());

What will be printed?

medium

A. 2

B. 1

C. 3

D. undefined

Solution

1. Step 1: Understand push operation

   Push adds element at end, then rotates queue to move new element to front, simulating stack top.

2. Step 2: Trace push and pop calls

   Push 1: queue=[1]; Push 2: queue=[2,1]; Push 3: queue=[3,2,1]; Pop removes front, which is 3.

3. Final Answer:

   3 -> Option C

4. Quick Check:

   Stack pop = last pushed = 3 [OK]

Hint: Rotate queue after push to keep newest element front [OK]

Common Mistakes:

• Assuming pop removes last element of array instead of front
• Not rotating queue after push
• Confusing queue shift() with stack pop()

4. The following code attempts to implement a stack using two queues. What is the error?

class Stack {
  constructor() {
    this.q1 = [];
    this.q2 = [];
  }
  push(x) {
    this.q1.push(x);
  }
  pop() {
    while (this.q1.length > 1) {
      this.q2.push(this.q1.shift());
    }
    return this.q1.shift();
  }
}

medium

A. pop() does not swap q1 and q2 after operation

B. push() should use q2 instead of q1

C. pop() should return q2.shift() instead of q1.shift()

D. Constructor should initialize q1 and q2 as objects, not arrays

Solution

1. Step 1: Analyze pop method

   Pop moves all but last element from q1 to q2, then removes last from q1.

2. Step 2: Identify missing step

   After pop, q1 is empty and q2 has elements, but q1 and q2 are not swapped, so next pop fails.

3. Final Answer:

   pop() does not swap q1 and q2 after operation -> Option A

4. Quick Check:

   Swap queues after pop to maintain stack [OK]

Hint: Always swap queues after pop in two-queue stack [OK]

Common Mistakes:

• Forgetting to swap queues after pop
• Returning from wrong queue
• Incorrect initialization of queues

5. You want to implement a stack using a single queue in JavaScript. Which approach correctly maintains stack order after each push operation?

class Stack {
  constructor() {
    this.queue = [];
  }
  push(x) {
    this.queue.push(x);
    // What should be done here?
  }
  pop() {
    return this.queue.shift();
  }
}

hard

A. Rotate the queue by dequeuing and enqueuing all elements except the last pushed

B. Sort the queue in descending order after each push

C. Clear the queue and add elements again in reverse order

D. Do nothing; queue order naturally matches stack order

Solution

1. Step 1: Understand stack order with one queue

   Stack is LIFO, but queue is FIFO, so after push, rotate queue to bring new element front.

2. Step 2: Rotate queue after push

   Remove elements from front and add them to back except the last pushed, so last pushed is at front.

3. Final Answer:

   Rotate the queue by dequeuing and enqueuing all elements except the last pushed -> Option A

4. Quick Check:

   Rotate queue after push to simulate stack [OK]

Hint: Rotate queue after push to keep newest element at front [OK]

Common Mistakes:

• Not rotating queue after push
• Trying to sort queue which is inefficient
• Assuming queue order matches stack without changes