Median of Data Stream Using Two Heaps in DSA C - Build from Scratch

Median of Data Stream Using Two Heaps

📖 Scenario: Imagine you are building a system that receives numbers one by one, like scores from a game played live. You want to find the middle score (median) quickly after each new score arrives.

🎯 Goal: Build a program in C that uses two heaps (a max heap and a min heap) to keep track of the median of a stream of numbers efficiently.

📋 What You'll Learn

Create two heaps: a max heap for the lower half of numbers and a min heap for the upper half.

Add numbers one by one to the heaps while keeping them balanced.

Calculate the median after each insertion.

Print the median after all numbers are added.

💡 Why This Matters

🌍 Real World

Finding median in real-time data streams like live sensor data, stock prices, or game scores.

💼 Career

Useful for software engineers working on real-time analytics, streaming data processing, and performance-critical applications.

1

Create arrays for two heaps

Create two integer arrays called maxHeap and minHeap each with size 100 to store heap elements. Also create two integer variables maxHeapSize and minHeapSize and set them to 0.

DSA C

# Create arrays maxHeap and minHeap with size 100
# Create integers maxHeapSize and minHeapSize and set to 0
// Your code here

Hint

Use int maxHeap[100]; and int minHeap[100]; to create arrays. Initialize sizes with int maxHeapSize = 0; and int minHeapSize = 0;.

2

Add helper functions for heap operations

Add function prototypes for void maxHeapInsert(int val), void minHeapInsert(int val), int maxHeapExtract(), and int minHeapExtract().

DSA C

#include 

int maxHeap[100];
int minHeap[100];
int maxHeapSize = 0;
int minHeapSize = 0;

// Declare heap operation functions
// Your code here

Hint

Write function declarations for the four heap operations with exact names and signatures.

3

Implement main logic to add numbers and balance heaps

Write a main function that adds these numbers in order: 5, 15, 1, 3 using maxHeapInsert and minHeapInsert. Balance the heaps so that maxHeapSize and minHeapSize differ by at most 1. Use the rule: if new number is less or equal to maxHeap root, insert into maxHeap; else insert into minHeap. After insertion, rebalance by moving elements between heaps if needed.

DSA C

#include 

int maxHeap[100];
int minHeap[100];
int maxHeapSize = 0;
int minHeapSize = 0;

void maxHeapInsert(int val);
void minHeapInsert(int val);
int maxHeapExtract();
int minHeapExtract();

int main() {
    // Insert numbers 5, 15, 1, 3 and balance heaps
    // Your code here
    return 0;
}

Hint

Use a loop to add numbers. Insert into maxHeap if number is smaller or equal to maxHeap root, else minHeap. Then rebalance heaps if size difference is more than 1.

4

Print the median after all insertions

In the main function, after adding and balancing heaps, print the median. If maxHeapSize is greater, median is maxHeap[0]. If minHeapSize is greater, median is minHeap[0]. If equal, median is average of maxHeap[0] and minHeap[0]. Use printf to print the median as a float with one decimal place.

DSA C

Hint

Check which heap is bigger to find median. If equal size, average the roots. Use printf("%.1f\n", median); to print.

Practice

1. What is the main purpose of using two heaps (a max-heap and a min-heap) to find the median of a data stream?

easy

A. To store all numbers in a single data structure

B. To sort all numbers in ascending order

C. To remove duplicate numbers from the stream

D. To efficiently keep track of the lower half and upper half of the numbers separately

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 this helps find the median

   By balancing these heaps, the median is either the top of one heap or the average of the tops of both heaps.

3. Final Answer:

   To efficiently keep track of the lower half and upper half of the numbers separately -> Option D

4. Quick Check:

   Two heaps split data halves = To efficiently keep track of the lower half and upper half of the numbers separately [OK]

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

Common Mistakes:

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

2. Which of the following is the correct way to insert a new number into the two heaps when maintaining the median?

easy

A. Insert into the heap based on comparison with max-heap top

B. Insert into min-heap only

C. Insert into max-heap only

D. Always insert into min-heap first, then balance heaps

Solution

1. Step 1: Decide where to insert the new number

   If the new number is less than or equal to the max-heap top, insert into max-heap; otherwise, insert into min-heap.

2. Step 2: Balance the heaps if needed

   After insertion, balance the heaps so their sizes differ by at most one.

3. Final Answer:

   Insert into the heap based on comparison with max-heap top -> Option A

4. Quick Check:

   Compare with max-heap top to choose heap [OK]

Hint: Compare new number with max-heap top to choose heap [OK]

Common Mistakes:

• Always inserting into one heap without condition
• Not balancing heaps after insertion
• Inserting into wrong heap based on min-heap top

3. Consider the following sequence of numbers inserted into the data stream: 5, 15, 1, 3. What is the median after all insertions using two heaps?

Insert 5
Insert 15
Insert 1
Insert 3
Find median

medium

A. 5

B. 3

C. 4

D. 6

Solution

1. Step 1: Insert numbers and balance heaps

   Insert 5 into max-heap ([5]); insert 15 into min-heap ([15]); insert 1 (<=5) into max-heap ([5,1] top=5); insert 3 (<=5) into max-heap ([5,3,1] top=5), sizes 3 vs 1, balance: min.push(5) ([15,5] top=5), max.pop() ([3,1] top=3).

2. Step 2: Calculate median from heaps

   Heaps sizes equal (2 and 2), median = average of max-heap top (3) and min-heap top (5) = (3+5)/2 = 4.

3. Final Answer:

   4 -> Option C

4. Quick Check:

   Median of [1,3,5,15] = 4 [OK]

Hint: Median is average when heaps equal size [OK]

Common Mistakes:

• Taking median as max-heap top only
• Not balancing heaps after insertions
• Calculating median incorrectly as 5 or 3

4. The following code snippet tries to maintain median using two heaps but has a bug. What is the bug?

void addNumber(int num, priority_queue<int>& maxHeap, priority_queue& minHeap) {
    if (maxHeap.empty() || num <= maxHeap.top()) {
        maxHeap.push(num);
    } else {
        minHeap.push(num);
    }
    if (maxHeap.size() > minHeap.size() + 1) {
        minHeap.push(maxHeap.top());
        maxHeap.pop();
    } else if (minHeap.size() > maxHeap.size() + 1) {
        maxHeap.push(minHeap.top());
        minHeap.pop();
    }
}

medium

A. The balancing pops top element before pushing, causing wrong order

B. The condition to insert into maxHeap is incorrect

C. The heaps are not balanced after insertion

D. The function does not return the median

Solution

1. Step 1: Analyze balancing logic

   When balancing, the code pops the top element of one heap before pushing the new top to the other, which moves the wrong element.

2. Step 2: Explain correct balancing

   Correct approach is to push heap.top() first, then pop() to transfer the correct top element.

3. Final Answer:

   The balancing pops top element before pushing, causing wrong order -> Option A

4. Quick Check:

   Push top before pop to balance heaps correctly [OK]

Hint: Push top before pop when balancing heaps [OK]

Common Mistakes:

• Popping before pushing the correct top
• Ignoring heap size difference conditions
• Not handling empty heaps properly

5. You have a data stream of integers and want to find the median after each insertion. Using two heaps, how would you handle the case when the total number of elements is odd?

hard

A. The median is always the top of the min-heap

B. The median is the top element of the heap with one extra element

C. The median is always the top of the max-heap

D. The median is the average of the tops of both heaps

Solution

1. Step 1: Understand heap size difference

   When total elements are odd, one heap will have one more element than the other.

2. Step 2: Determine median from heaps

   The median is the top element of the heap that has the extra element, because it holds the middle value.

3. Final Answer:

   The median is the top element of the heap with one extra element -> Option B

4. Quick Check:

   Odd count median = top of larger heap [OK]

Hint: Median is top of bigger heap when odd count [OK]

Common Mistakes:

• Averaging tops when count is odd
• Assuming max-heap always has extra element
• Ignoring heap size difference