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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to add a number to the max heap.

DSA Typescript

maxHeap.[1](num);

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

Apop

Bpeek

Csort

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push to add elements.

Trying to use peek which only views the top element.

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 size is greater than min heap size by more than one.

DSA Typescript

minHeap.[1](maxHeap.[2]());

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

Ashift

Bpop

Cpeek

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek instead of pop to remove element.

Using shift which is not a heap method.

We pop the top from maxHeap and push it into minHeap to balance sizes.

3fill in blank

hard

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

DSA Typescript

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

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

Apeek

Bpop

Cpush

Dsort

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() removes elements and changes the heap.

Using push() is for adding elements, not retrieving.

We use peek() to get the top elements without removing them.

4fill in blank

hard

Fill both blanks to correctly insert a number into the appropriate heap.

DSA Typescript

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

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

A<=

B>

Cpeek

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop() instead of peek() to compare.

Using > instead of <= changes the logic.

We compare num with the top of maxHeap using peek() and insert accordingly.

5fill in blank

hard

Fill all three blanks to rebalance heaps when min heap size is greater than max heap size.

DSA Typescript

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

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

A1

Bpop

Cpush

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using 2 instead of 1 for size difference check.

Using pop and push in wrong order.

We check if minHeap is bigger by 1, then push the popped element from minHeap to maxHeap.