Median of Data Stream Using Two Heaps in DSA Java - Time & Space Complexity

Time Complexity: Median of Data Stream Using Two Heaps

O(log n)

Understanding Time Complexity

We want to understand how fast we can find the middle value when numbers keep coming in.

How does the time to add a number and get the middle change as more numbers arrive?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

import java.util.PriorityQueue;

class MedianFinder {
    PriorityQueue<Integer> lowers = new PriorityQueue<>((a, b) -> b - a); // max heap
    PriorityQueue<Integer> highers = new PriorityQueue<>(); // min heap

    public void addNum(int num) {
        if (lowers.isEmpty() || num <= lowers.peek()) lowers.add(num);
        else highers.add(num);

        if (lowers.size() - highers.size() > 1) highers.add(lowers.poll());
        else if (highers.size() - lowers.size() > 1) lowers.add(highers.poll());
    }

    public double findMedian() {
        if (lowers.size() > highers.size()) return lowers.peek();
        else if (highers.size() > lowers.size()) return highers.peek();
        else return (lowers.peek() + highers.peek()) / 2.0;
    }
}
    

This code keeps two heaps to quickly find the median as numbers are added.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Adding a number to one of the two heaps and balancing them.
• How many times: Each addNum call does a few heap operations, each taking time related to heap size.

How Execution Grows With Input

Each new number is added to a heap and sometimes causes a rebalance between heaps.

Input Size (n)	Approx. Operations
10	About 10 insertions and possible rebalances, each small and fast.
100	About 100 insertions, each heap operation slightly slower but still quick.
1000	About 1000 insertions, heap operations take a bit longer but grow slowly.

Pattern observation: The time per insertion grows slowly as the heaps get bigger, not linearly.

Final Time Complexity

Time Complexity: O(log n)

This means adding each number and updating the median takes time that grows slowly with the total numbers seen.

Common Mistake

[X] Wrong: "Adding a number is constant time because it's just one insert."

[OK] Correct: Inserting into a heap requires rearranging elements to keep order, which takes time growing with the heap size.

Interview Connect

Knowing how to keep track of the middle value quickly is a useful skill for many real-time data problems.

Self-Check

"What if we used a sorted list instead of two heaps? How would the time complexity change?"

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