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

Time Complexity: Median of Data Stream Using Two Heaps

O(log n)

Understanding Time Complexity

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

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

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class MedianFinder {
  constructor() {
    this.small = new MaxHeap(); // lower half
    this.large = new MinHeap(); // upper half
  }

  addNum(num) {
    this.small.insert(num);
    this.large.insert(this.small.extractMax());
    if (this.large.size() > this.small.size()) {
      this.small.insert(this.large.extractMin());
    }
  }

  findMedian() {
    if (this.small.size() > this.large.size()) return this.small.peek();
    return (this.small.peek() + this.large.peek()) / 2;
  }
}
    

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

Identify Repeating Operations

Look at what repeats when adding numbers.

• Primary operation: Inserting and extracting from heaps (insert, extractMax, extractMin).
• How many times: Each addNum call does a few heap operations, each on a smaller set of numbers.

How Execution Grows With Input

Each new number causes a few heap operations that take longer as the heaps grow.

Input Size (n)	Approx. Operations
10	About 10 insertions and extractions, each quick.
100	About 100 insertions and extractions, each a bit slower.
1000	About 1000 insertions and extractions, each slower but still efficient.

Pattern observation: The time per add grows slowly, roughly like the height of the heaps, which grows with the log of input size.

Final Time Complexity

Time Complexity: O(log n)

This means adding each number and updating the median takes time that grows slowly as more numbers come in.

Common Mistake

[X] Wrong: "Adding a number takes constant time because we just put it somewhere."

[OK] Correct: Actually, we must keep the heaps balanced and ordered, so inserting and extracting take time that grows with the heap size.

Interview Connect

Understanding this method shows you can handle data that changes over time and still find answers quickly.

Self-Check

"What if we used only one heap instead of two? How would the time complexity change?"

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