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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the max heap for the lower half of numbers.

DSA Java

PriorityQueue<Integer> maxHeap = new PriorityQueue<>([1]);

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

A() -> {}

B(a, b) -> b - a

C(a, b) -> a - b

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using natural order comparator which creates a min heap instead of max heap.

Passing null which creates a min heap by default.

The max heap is created by providing a comparator that reverses the natural order, so the largest element is at the top.

2fill in blank

medium

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

DSA Java

maxHeap.[1](num);

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

Aoffer

Bpop

Cpeek

Dpoll

Attempts:

3 left

💡 Hint

Common Mistakes

Using poll or pop which remove elements instead of adding.

Using peek which only returns the top element without removing or adding.

The offer method adds an element to the PriorityQueue.

3fill in blank

hard

Fix the error in balancing heaps by completing the condition to check if max heap size is greater than min heap size by more than one.

DSA Java

if (maxHeap.size() [1] minHeap.size() + 1) {

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

A<

B==

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than operator which reverses the logic.

Using equality operator which only triggers when sizes are equal.

We check if max heap size is greater than min heap size by more than one to rebalance.

4fill in blank

hard

Fill both blanks to complete the code that moves the top element from max heap to min heap.

DSA Java

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

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

Aoffer

Bpoll

Cpeek

Dadd

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek which does not remove the element.

Using add instead of offer which is preferred for PriorityQueue.

We remove the top element from max heap using poll and add it to min heap using offer.

5fill in blank

hard

Fill all three blanks to complete the code that returns the median from the two heaps.

DSA Java

if (maxHeap.size() == minHeap.size()) {
    return (maxHeap.[1]() + minHeap.[2]()) / 2.0;
} else {
    return maxHeap.[3]();
}

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

Apeek

Bpoll

Csize

Delement

Attempts:

3 left

💡 Hint

Common Mistakes

Using poll which removes elements and changes heap state.

Using size which returns the number of elements, not the top element.

We use peek to look at the top elements without removing them to calculate the median.

Practice

1. What is the main purpose of using two heaps (a max heap and a min heap) in the Median of Data Stream problem?

easy

A. To store all numbers in a single data structure

B. To sort all numbers in ascending order

C. To efficiently keep track of the lower half and upper half of the numbers for median calculation

D. To find the maximum number in the data stream

Solution

1. Step 1: Understand the role of two heaps

   The max heap stores the smaller half of the numbers, and the min heap stores the larger half.

2. Step 2: Explain median calculation

   By balancing these heaps, the median can be found quickly from the top elements of the heaps.

3. Final Answer:

   To efficiently keep track of the lower half and upper half of the numbers for median calculation -> Option C

4. Quick Check:

   Two heaps split data for median [OK]

Hint: Two heaps split data into halves for quick median [OK]

Common Mistakes:

• Thinking heaps sort all numbers completely
• Using only one heap for median
• Confusing max heap and min heap roles

2. Which of the following Java code snippets correctly initializes a max heap using PriorityQueue for the Median of Data Stream problem?

easy

A. PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());

B. PriorityQueue<Integer> maxHeap = new PriorityQueue<>();

C. PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Comparator.naturalOrder());

D. PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.naturalOrder());

Solution

1. Step 1: Identify max heap initialization

   Java's PriorityQueue is a min heap by default; to make max heap, use Collections.reverseOrder().

2. Step 2: Check options

   PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder()); uses Collections.reverseOrder(), which correctly creates a max heap.

3. Final Answer:

   PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder()); -> Option A

4. Quick Check:

   Max heap needs reverse order comparator [OK]

Hint: Use Collections.reverseOrder() for max heap in Java [OK]

Common Mistakes:

• Using default PriorityQueue without comparator
• Using naturalOrder() which is min heap
• Using Collections.naturalOrder() which doesn't exist

3. Consider the following Java code snippet for adding numbers to two heaps and finding the median:

MedianFinder mf = new MedianFinder();
mf.addNum(1);
mf.addNum(5);
mf.addNum(2);
System.out.println(mf.findMedian());

Assuming the MedianFinder class uses two heaps correctly, what will be the printed output?

medium

A. 2.5

B. 2

C. 3

D. 1

Solution

1. Step 1: Add numbers and balance heaps

   Numbers added: 1, 5, 2. Sorted order: 1, 2, 5. Median is middle number 2.

2. Step 2: Calculate median

   Since odd count, median is the root of the heap with one extra element, which is 2.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Median of [1,2,5] is 2 [OK]

Hint: Median of odd count is middle element [OK]

Common Mistakes:

• Calculating average instead of middle for odd count
• Ignoring heap balancing
• Confusing median with mean

4. The following Java method is part of a MedianFinder class using two heaps. It is supposed to add a number and rebalance heaps. Identify the bug:

public void addNum(int num) {
    if (maxHeap.isEmpty() || num <= maxHeap.peek()) {
        maxHeap.add(num);
    } else {
        minHeap.add(num);
    }
    if (maxHeap.size() > minHeap.size() + 1) {
        minHeap.add(maxHeap.poll());
    } else if (minHeap.size() > maxHeap.size()) {
        maxHeap.add(minHeap.poll());
    }
}

medium

A. Should check if maxHeap is empty before peek() to avoid NullPointerException

B. The addNum method should not add to minHeap when num equals maxHeap.peek()

C. The rebalance condition for minHeap size is incorrect; should be minHeap.size() > maxHeap.size() + 1

D. No bug, method works correctly

Solution

1. Step 1: Analyze addNum logic

   It adds to maxHeap if empty or num <= maxHeap.peek(), else to minHeap.

2. Step 2: Check rebalance conditions

   Rebalances if maxHeap is bigger by more than 1 or minHeap bigger than maxHeap.

3. Final Answer:

   No bug, method works correctly -> Option D

4. Quick Check:

   Correct heap balancing logic [OK]

Hint: Check heap sizes differ by at most one [OK]

Common Mistakes:

• Forgetting to check maxHeap empty before peek
• Incorrect rebalance size conditions
• Misplacing elements between heaps

5. You want to extend the MedianFinder class to handle a large data stream efficiently. Which approach improves performance when finding the median after every insertion?

hard

A. Use two balanced heaps to keep halves and rebalance after each insertion

B. Store all numbers in a list and sort it after each insertion

C. Use a single max heap to store all numbers

D. Use a hash map to count frequencies and calculate median from counts

Solution

1. Step 1: Evaluate approaches for median calculation

   Sorting after each insertion (Store all numbers in a list and sort it after each insertion) is costly (O(n log n)). Single heap or hash map do not support quick median.

2. Step 2: Explain two heaps efficiency

   Two heaps keep halves balanced, allowing O(log n) insertion and O(1) median retrieval.

3. Final Answer:

   Use two balanced heaps to keep halves and rebalance after each insertion -> Option A

4. Quick Check:

   Two heaps optimize median in data stream [OK]

Hint: Two heaps give fast median updates in streams [OK]

Common Mistakes:

• Sorting entire list each time
• Using single heap only
• Ignoring insertion cost