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

Median of Data Stream Using Two Heaps

📖 Scenario: Imagine you are building a live dashboard that shows the median value of numbers coming in one by one, like temperatures recorded every minute. You want to keep track of the median efficiently as new numbers arrive.

🎯 Goal: Build a program that uses two heaps (a max heap and a min heap) to find the median of a stream of numbers step-by-step.

📋 What You'll Learn

Create two heaps: a max heap for the smaller half of numbers and a min heap for the larger half.

Add numbers one by one to the correct heap.

Balance the heaps so their sizes differ by at most one.

Calculate the median from the heaps after all numbers are added.

💡 Why This Matters

🌍 Real World

Finding median in real-time data streams like sensor readings, stock prices, or live user inputs.

💼 Career

Useful for roles in data engineering, software development, and analytics where streaming data needs quick summary statistics.

1

Create two empty heaps

Create two empty arrays called maxHeap and minHeap to represent the max heap and min heap respectively.

DSA Javascript

# Your code here
const maxHeap = [];
const minHeap = [];

Hint

Use empty arrays to start your heaps.

2

Add a helper function to insert into max heap

Write a function called insertMaxHeap that takes a number and inserts it into maxHeap while keeping it sorted in descending order.

DSA Javascript

const maxHeap = [];
const minHeap = [];

// Write function insertMaxHeap to insert number in descending order
// Your code here

Hint

Push the number then sort descending to simulate max heap.

3

Add a helper function to insert into min heap

Write a function called insertMinHeap that takes a number and inserts it into minHeap while keeping it sorted in ascending order.

DSA Javascript

const maxHeap = [];
const minHeap = [];

function insertMaxHeap(num) {
    maxHeap.push(num);
    maxHeap.sort((a, b) => b - a);
}

// Write function insertMinHeap to insert number in ascending order
// Your code here

Hint

Push the number then sort ascending to simulate min heap.

4

Add numbers and find median

Write code to add these numbers 5, 15, 1, 3 one by one to the heaps using insertMaxHeap and insertMinHeap. Keep the heaps balanced so their sizes differ by at most one. Then calculate and print the median after all insertions.

DSA Javascript

const maxHeap = [];
const minHeap = [];

function insertMaxHeap(num) {
    maxHeap.push(num);
    maxHeap.sort((a, b) => b - a);
}

function insertMinHeap(num) {
    minHeap.push(num);
    minHeap.sort((a, b) => a - b);
}

// Add numbers 5, 15, 1, 3 to heaps and balance them
// Calculate and print the median
// Your code here

Hint

Insert each number, balance heaps, then calculate median as middle value or average of two middles.

Practice

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

easy

A. To sort all numbers in ascending order

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

C. To store all numbers in a single data structure

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 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 B

4. Quick Check:

   Two heaps split data halves = A [OK]

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

Common Mistakes:

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

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

easy

A. Always insert into min-heap

B. Always insert into max-heap

C. Insert into max-heap if number is less than or equal to max-heap root; else insert into min-heap

D. Insert randomly into either heap

Solution

1. Step 1: Decide heap based on number value

   If the number is smaller or equal to the max-heap root, it belongs to the lower half, so insert into max-heap.

2. Step 2: Otherwise insert into min-heap

   If the number is greater, it belongs to the upper half, so insert into min-heap.

3. Final Answer:

   Insert into max-heap if number is less than or equal to max-heap root; else insert into min-heap -> Option C

4. Quick Check:

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

Hint: Compare with max-heap root to choose heap [OK]

Common Mistakes:

• Inserting all numbers into one heap
• Not comparing with max-heap root
• Inserting randomly without condition

3. Consider the following JavaScript code snippet using two heaps to find the median after inserting numbers 5, 15, 1, 3:

const maxHeap = new MaxHeap();
const minHeap = new MinHeap();

function addNum(num) {
  if (maxHeap.size() === 0 || num <= maxHeap.peek()) {
    maxHeap.insert(num);
  } else {
    minHeap.insert(num);
  }

  if (maxHeap.size() > minHeap.size() + 1) {
    minHeap.insert(maxHeap.extract());
  } else if (minHeap.size() > maxHeap.size()) {
    maxHeap.insert(minHeap.extract());
  }
}

addNum(5);
addNum(15);
addNum(1);
addNum(3);

const median = maxHeap.size() > minHeap.size() ? maxHeap.peek() : (maxHeap.peek() + minHeap.peek()) / 2;
console.log(median);

What is the output printed?

medium

A. 4

B. 3

C. 5

D. 6

Solution

1. Step 1: Insert numbers and balance heaps

   Insert 5 into maxHeap (empty). Insert 15 into minHeap (15 > 5). Insert 1 into maxHeap (1 <= 5). Insert 3 into maxHeap (3 <= 5). Now maxHeap has 3 elements, minHeap has 1.

2. Step 2: Balance heaps sizes

   maxHeap size (3) > minHeap size (1) + 1, so move maxHeap.extract() (which is 5) to minHeap. Now maxHeap has [3,1], minHeap has [5,15]. Median is average of maxHeap.peek()=3 and minHeap.peek()=5, so (3+5)/2 = 4.

3. Final Answer:

   4 -> Option A

4. Quick Check:

   Median after balancing = 4 [OK]

Hint: Balance heaps sizes, median is middle or average [OK]

Common Mistakes:

• Not balancing heaps after insertion
• Taking median as maxHeap.peek() only
• Incorrect heap extraction order

4. The following code snippet is intended to maintain the median using two heaps, but it has a bug:

function addNum(num) {
  if (num < maxHeap.peek()) {
    maxHeap.insert(num);
  } else {
    minHeap.insert(num);
  }

  if (maxHeap.size() > minHeap.size() + 1) {
    minHeap.insert(maxHeap.extract());
  } else if (minHeap.size() > maxHeap.size()) {
    maxHeap.insert(minHeap.extract());
  }
}

What is the bug in this code?

medium

A. It inserts all numbers into minHeap

B. It uses > instead of >= in the size comparison

C. It balances heaps incorrectly by moving elements the wrong way

D. It does not handle the case when maxHeap is empty before peek()

Solution

1. Step 1: Identify peek() usage problem

   The code calls maxHeap.peek() without checking if maxHeap is empty, causing an error on first insertion.

2. Step 2: Explain impact of missing empty check

   Without checking, calling peek() on empty maxHeap throws an error, so the first number cannot be inserted properly.

3. Final Answer:

   It does not handle the case when maxHeap is empty before peek() -> Option D

4. Quick Check:

   Empty maxHeap peek() causes error [OK]

Hint: Check heap empty before peek() to avoid errors [OK]

Common Mistakes:

• Ignoring empty heap before peek()
• Assuming heaps always balanced before insert
• Incorrect size comparison operators

5. You are given a stream of numbers: [2, 4, 7, 1, 5, 3]. Using two heaps (max-heap for lower half, min-heap for upper half), what is the median after inserting all numbers? Also, explain why balancing the heaps is important in this process.

hard

A. 3.5; Balancing keeps heaps sizes equal or off by one to find median correctly

B. 4; Balancing is not necessary if we always insert correctly

C. 3; Balancing helps to sort the entire stream

D. 5; Balancing is used to find max value quickly

Solution

1. Step 1: Insert numbers and balance heaps

   Insert 2 (maxHeap empty, goes to maxHeap). Insert 4 (4 > 2, minHeap). Insert 7 (7 > 2, minHeap). Now minHeap has 2 elements, maxHeap 1, so move minHeap.extract() (4) to maxHeap. Insert 1 (1 <= 4, maxHeap). Insert 5 (5 > 4, minHeap). Insert 3 (3 <= 4, maxHeap). Heaps are balanced with maxHeap size 4 and minHeap size 2, move maxHeap.extract() (4) to minHeap to balance sizes.

2. Step 2: Calculate median and explain balancing

   Now maxHeap has [3,2,1] with peek()=3, minHeap has [4,5,7] with peek()=4. Since sizes equal (3 and 3), median is (3 + 4)/2 = 3.5. Balancing keeps heaps sizes equal or off by one to find median correctly from the tops.

3. Final Answer:

   3.5; Balancing keeps heaps sizes equal or off by one to find median correctly -> Option A

4. Quick Check:

   Balanced heaps give median 3.5 [OK]

Hint: Balance heaps sizes to get median from tops [OK]

Common Mistakes:

• Not balancing heaps after each insertion
• Calculating median from unsorted data
• Confusing median with max or min values