Why Median of Data Stream Using Two Heaps in DSA C++?

The Big Idea

What if you could instantly know the middle value of a never-ending list without sorting it all over and over?

The Scenario

Imagine you are tracking the middle value of a long list of numbers that keep coming in one by one, like scores in a game or temperatures throughout the day.

You try to write down all numbers and sort them every time a new number arrives to find the middle one.

The Problem

Sorting the entire list every time a new number arrives is very slow and tiring, especially if the list grows very large.

It also wastes a lot of time and effort, making it hard to get the middle value quickly.

The Solution

Using two heaps (special trees that keep numbers in order), one to keep the smaller half and one for the larger half, lets you quickly find the middle number without sorting everything again.

This way, you only adjust a few numbers each time, making the process fast and easy.

Before vs After

✗ Before

vector<int> numbers;
// On new number:
numbers.push_back(new_number);
sort(numbers.begin(), numbers.end());
int median = numbers[numbers.size() / 2];

✓ After

priority_queue<int> maxHeap; // smaller half
priority_queue<int, vector<int>, greater<int>> minHeap; // larger half
// On new number:
if (maxHeap.empty() || new_number <= maxHeap.top()) maxHeap.push(new_number);
else minHeap.push(new_number);
// Balance heaps
if (maxHeap.size() > minHeap.size() + 1) {
    minHeap.push(maxHeap.top());
    maxHeap.pop();
} else if (minHeap.size() > maxHeap.size()) {
    maxHeap.push(minHeap.top());
    minHeap.pop();
}
// Median is top of maxHeap or average of tops

What It Enables

This method makes it possible to find the middle value instantly as new numbers keep coming in, no matter how many numbers there are.

Real Life Example

Streaming services use this to show the median rating of a movie as users keep rating it live, without waiting for all ratings to finish.

Key Takeaways

Manually sorting every time is slow and inefficient.

Two heaps keep track of smaller and larger halves efficiently.

Median can be found quickly as data streams in.

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 efficiently find the median by balancing smaller and larger halves of the data

B. To sort the entire data stream in ascending order

C. To store all elements in one heap for quick access

D. To remove duplicate elements from 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 how median is found

   Balancing these heaps allows quick access to the middle values, enabling efficient median calculation.

3. Final Answer:

   To efficiently find the median by balancing smaller and larger halves of the data -> Option A

4. Quick Check:

   Two heaps balance halves = median found fast [OK]

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

Common Mistakes:

• Thinking heaps sort entire data
• Using one heap only
• Confusing heaps with arrays

2. Which of the following is the default way to declare a max-heap in C++ using the priority_queue?

easy

A. priority_queue maxHeap;

B. priority_queue<int> maxHeap;

C. priority_queue maxHeap;

D. priority_queue maxHeap;

Solution

1. Step 1: Recall default priority_queue behavior

   By default, priority_queue in C++ is a max-heap using less<int>.

2. Step 2: Identify correct declaration

   Declaring priority_queue<int> maxHeap; creates a max-heap without extra parameters.

3. Final Answer:

   priority_queue<int> maxHeap; -> Option B

4. Quick Check:

   Default priority_queue = max-heap [OK]

Hint: Default priority_queue is max-heap [OK]

Common Mistakes:

• Using greater<int> for max-heap
• Omitting vector container
• Confusing max-heap with min-heap syntax

3. Consider the following C++ code snippet that inserts numbers into two heaps to find the median:

priority_queue<int> maxHeap;
priority_queue minHeap;

// Insert 10
maxHeap.push(10);
// Insert 20
minHeap.push(20);
// Insert 15
if (15 <= maxHeap.top()) maxHeap.push(15); else minHeap.push(15);

// Balance heaps
if (maxHeap.size() > minHeap.size() + 1) {
  minHeap.push(maxHeap.top());
  maxHeap.pop();
} else if (minHeap.size() > maxHeap.size()) {
  maxHeap.push(minHeap.top());
  minHeap.pop();
}

// Calculate median
if (maxHeap.size() == minHeap.size())
  cout << (maxHeap.top() + minHeap.top()) / 2.0 << endl;
else
  cout << maxHeap.top() << endl;

What is the output of this code?

medium

A. 15

B. 15.0

C. 12.5

D. 20

Solution

1. Step 1: Insert elements and balance heaps

   Insert 10 into maxHeap: maxHeap = [10]. Insert 20 into minHeap: minHeap = [20]. Insert 15: 15 > 10, so goes to minHeap: minHeap = [15, 20]. Balance heaps: minHeap size (2) > maxHeap size (1), so move minHeap top (15) to maxHeap. Now maxHeap = [15, 10], minHeap = [20].

2. Step 2: Calculate median

   maxHeap size = 2, minHeap size = 1, so median is maxHeap.top() = 15. Sizes unequal, cout << maxHeap.top() << endl; outputs 15.

3. Final Answer:

   15 -> Option A

4. Quick Check:

   Median after balancing = 15 [OK]

Hint: Balance heaps, median = top of larger heap or average [OK]

Common Mistakes:

• Not balancing heaps correctly
• Integer division instead of float
• Wrong heap insertion condition

4. The following code snippet is intended to maintain two heaps for median calculation. Identify the bug:

priority_queue<int> maxHeap;
priority_queue minHeap;

void addNumber(int num) {
  if (maxHeap.empty() || num <= maxHeap.top()) {
    maxHeap.push(num);
  } else {
    minHeap.push(num);
  }

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

medium

A. The heaps are not declared correctly for median calculation

B. The comparison operator in if condition should be num >= maxHeap.top()

C. The balancing condition for maxHeap is incorrect; it should check if maxHeap.size() > minHeap.size() + 1

D. The function does not insert elements into any heap

Solution

1. Step 1: Analyze balancing conditions

   The code moves element from maxHeap to minHeap if maxHeap.size() > minHeap.size(), but it should only do so if maxHeap.size() > minHeap.size() + 1 to keep heaps balanced properly.

2. Step 2: Confirm correct balancing logic

   Proper balancing ensures size difference is at most 1. The current condition causes unnecessary moves, breaking balance.

3. Final Answer:

   The balancing condition for maxHeap is incorrect; it should check if maxHeap.size() > minHeap.size() + 1 -> Option C

4. Quick Check:

   Balance heaps only if size difference > 1 [OK]

Hint: Balance heaps only if size difference > 1 [OK]

Common Mistakes:

• Wrong balancing condition
• Swapping heap roles
• Incorrect comparison operators

5. You are given a stream of integers: [5, 15, 1, 3, 8, 7, 9, 10, 6, 11]. Using two heaps to find the median after each insertion, what is the median after inserting all elements?

hard

A. 8

B. 7

C. 6.5

D. 7.5

Solution

1. Step 1: Insert elements and balance heaps step-by-step

   After processing the entire stream with proper insertion (to maxHeap if <= max.top or empty, else minHeap) and balancing (rebalance if max.size > min.size + 1 or min.size > max.size), final state: maxHeap size=5 (top=7), minHeap size=5 (top=8).

2. Step 2: Calculate final median

   After all insertions, maxHeap size = 5, minHeap size = 5, median = average of tops: (7 + 8) / 2 = 7.5

3. Final Answer:

   7.5 -> Option D

4. Quick Check:

   Median after all insertions = 7.5 [OK]

Hint: Median = average of two heap tops if equal size [OK]

Common Mistakes:

• Not balancing heaps after each insertion
• Calculating median from wrong heap
• Forgetting to average when heaps equal size