Median of Data Stream Using Two Heaps in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a number into the max heap.

DSA Javascript

maxHeap.[1](num);

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

Apeek

Bpop

Csort

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push removes elements instead of adding.

Using peek only looks at the top element, does not add.

We use push to add a new number to the max heap.

2fill in blank

medium

Complete the code to balance the heaps when max heap has more elements.

DSA Javascript

if (maxHeap.size() > minHeap.size() + [1]) {
  minHeap.push(maxHeap.pop());
}

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

A1

B0

C2

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using 2 or more causes imbalance.

Using 0 means no difference allowed, which is too strict.

We balance heaps when max heap size is more than min heap size by 1.

3fill in blank

hard

Fix the error in the code to get the median when heaps are equal size.

DSA Javascript

return (maxHeap.peek() [1] minHeap.peek()) / 2;

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

A+

B-

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction or multiplication instead of addition.

Dividing before adding changes the result.

Median is average of top elements, so we add then divide by 2.

4fill in blank

hard

Fill both blanks to correctly decide which heap to push the new number into.

DSA Javascript

if (num [1] maxHeap.peek()) {
  maxHeap.push(num);
} else {
  minHeap.[2](num);
}

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

A<

B>

Cpush

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < reverses logic.

Using pop instead of push removes elements.

If the number is less than max heap top, push to max heap; else push to min heap.

5fill in blank

hard

Fill all three blanks to correctly rebalance heaps and return the median.

DSA Javascript

if (maxHeap.size() > minHeap.size() + [1]) {
  minHeap.push(maxHeap.[2]());
} else if (minHeap.size() > maxHeap.size() + [3]) {
  maxHeap.push(minHeap.pop());
}

if (maxHeap.size() === minHeap.size()) {
  return (maxHeap.peek() + minHeap.peek()) / 2;
} else {
  return maxHeap.peek();
}

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

A1

Bpop

C0

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 as size difference allows no difference, causing unnecessary moves.

Using push instead of pop to remove element causes errors.

We rebalance when size difference is more than 1 by moving top element using pop, then return median.