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

Practice - 5 Tasks

1fill in blank

easy

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

DSA C++

maxHeap.[1](num);

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

Aempty

Bpop

Ctop

Dpush

Attempts:

3 left

💡 Hint

Common Mistakes

Using pop instead of push to add elements.

Trying to access top instead of inserting.

We use push to add a new element to the heap.

2fill in blank

medium

Complete the code to balance the sizes of the two heaps after insertion.

DSA C++

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

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

A0

B2

C1

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or 2 instead of 1 for size difference.

Not balancing heaps after insertion.

We balance heaps when maxHeap size is more than minHeap size by 1.

3fill in blank

hard

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

DSA C++

return (maxHeap.top() + [1].top()) / 2.0;

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AminHeap

BmaxHeap

Cheap

DmedianHeap

Attempts:

3 left

💡 Hint

Common Mistakes

Using maxHeap.top() twice.

Using undefined heap names.

The median is average of tops of maxHeap and minHeap, so use minHeap.

4fill in blank

hard

Fill both blanks to insert a number into the correct heap based on its value.

DSA C++

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

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

A<

B>

Cpush

Dpop

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < for comparison.

Using pop instead of push to insert.

If number is less than maxHeap top, insert into maxHeap using push.

5fill in blank

hard

Fill all three blanks to return the median correctly based on heap sizes.

DSA C++

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

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

Atop

C2.0

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using size() instead of top() to get element.

Dividing by 2 instead of 2.0 causing integer division.

Return top of the bigger heap or average of both tops divided by 2.0.