Median of Data Stream Using Two Heaps in DSA Typescript - 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 one by one.

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.

class MedianFinder {
  private maxHeap: number[] = [] // smaller half
  private minHeap: number[] = [] // larger half

  addNum(num: number): void {
    if (!this.maxHeap.length || num <= -this.maxHeap[0]) {
      this.maxHeapPush(-num)
    } else {
      this.minHeapPush(num)
    }
    this.balanceHeaps()
  }

  findMedian(): number {
    if (this.maxHeap.length > this.minHeap.length) {
      return -this.maxHeap[0]
    } else {
      return (-this.maxHeap[0] + this.minHeap[0]) / 2
    }
  }

  private maxHeapPush(val: number) { /* push and heapify */ }
  private minHeapPush(val: number) { /* push and heapify */ }
  private balanceHeaps() { /* move elements to keep sizes balanced */ }
}
    

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 involves inserting into a heap and balancing heaps.
• How many times: Each addNum call does these operations once per number added.

How Execution Grows With Input

Each new number causes a few heap operations that take time related to the heap size.

Input Size (n)	Approx. Operations
10	About 10 insertions and balancing steps, each small.
100	About 100 insertions, each a bit longer due to bigger heaps.
1000	About 1000 insertions, each slightly longer but still efficient.

Pattern observation: The time per insertion grows slowly as heaps grow, 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 added.

Common Mistake

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

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

Interview Connect

Knowing how to keep track of medians efficiently shows you can handle data that changes over time, a useful skill in many real problems.

Self-Check

"What if we used a sorted array instead of 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 sort all numbers in ascending order

B. To store all numbers in a single data structure

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

D. To remove duplicates 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 this helps find the median

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

3. Final Answer:

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

4. Quick Check:

   Two heaps split data halves = efficient median [OK]

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

Common Mistakes:

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

2. Which TypeScript syntax correctly declares a max heap using a priority queue for numbers?

easy

A. const maxHeap = new PriorityQueue<number>((a, b) => b - a);

B. const maxHeap = new PriorityQueue<number>((a, b) => a - b);

C. const maxHeap = new PriorityQueue<number>();

D. const maxHeap = new PriorityQueue<number>((a, b) => a + b);

Solution

1. Step 1: Understand max heap comparator

   Max heap comparator returns positive if first is less than second, so b - a creates max heap.

2. Step 2: Check each option

   const maxHeap = new PriorityQueue<number>((a, b) => b - a); uses (b - a) which is correct for max heap; others do not create max heap order.

3. Final Answer:

   const maxHeap = new PriorityQueue<number>((a, b) => b - a); -> Option A

4. Quick Check:

   Comparator b - a = max heap [OK]

Hint: Max heap comparator subtracts first from second [OK]

Common Mistakes:

• Using a - b for max heap (creates min heap)
• Not providing comparator
• Using addition in comparator

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

medium

A. 3

B. 4

C. 5

D. 8

Solution

1. Step 1: Insert numbers and balance heaps

   Insert 5 -> maxHeap=[5] (top=5), minHeap=[]; Insert 15 -> minHeap=[15] (top=15), sizes 1:1; Insert 1 -> maxHeap=[5,1] (top=5), sizes 2:1; Insert 3 -> maxHeap=[5,3,1] (top=5), sizes 3:1 -> balance (max.size > min.size+1), move max.top=5 to minHeap -> maxHeap=[3,1] (top=3), minHeap=[5,15] (top=5), sizes 2:2.

2. Step 2: Calculate median from heaps

   Sizes equal, median = (maxHeap.top + minHeap.top)/2 = (3 + 5)/2 = 4.

3. Final Answer:

   4 -> Option B

4. Quick Check:

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

Hint: Median is average of two heap tops if sizes equal [OK]

Common Mistakes:

• Taking median as top of one heap only
• Not balancing heaps after insertion
• Wrong average calculation

4. Consider this TypeScript snippet for adding a number to the data stream median finder:

if (num <= maxHeap.peek()) {
  maxHeap.enqueue(num);
} else {
  minHeap.enqueue(num);
}

// Balance heaps
if (maxHeap.size() > minHeap.size() + 1) {
  minHeap.enqueue(maxHeap.dequeue());
} else if (minHeap.size() > maxHeap.size()) {
  maxHeap.enqueue(minHeap.dequeue());
}

What is the bug in this code?

medium

A. It compares num with minHeap.peek() instead of maxHeap.peek()

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

C. It uses enqueue instead of push for heaps

D. It does not handle empty maxHeap before peek() call

Solution

1. Step 1: Analyze peek() usage on maxHeap

   If maxHeap is empty, calling peek() causes error because no element exists yet.

2. Step 2: Identify missing empty check

   Code should check if maxHeap is empty before peek() or handle initial insertion separately.

3. Final Answer:

   It does not handle empty maxHeap before peek() call -> Option D

4. Quick Check:

   Empty heap peek() causes runtime error [OK]

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

Common Mistakes:

• Assuming heaps always have elements
• Mixing enqueue and push methods
• Balancing heaps incorrectly

5. You want to extend the Median of Data Stream solution to support removing numbers from the stream efficiently. Which data structure modification is best to support both insertion and removal while maintaining median calculation?

hard

A. Use two heaps with a hash map to mark removed elements lazily

B. Use two balanced binary search trees (like AVL or Red-Black trees) instead of heaps

C. Use a single sorted array and binary search for insertion and removal

D. Use two queues to store lower and upper halves

Solution

1. Step 1: Understand removal challenge in heaps

   Heaps do not support efficient arbitrary removal, so direct removal is costly.

2. Step 2: Explain lazy removal with hash map

   Mark elements as removed in a hash map and skip them during heap operations to simulate removal efficiently.

3. Step 3: Compare other options

   Balanced trees support removal but are more complex; arrays are inefficient; queues do not maintain order.

4. Final Answer:

   Use two heaps with a hash map to mark removed elements lazily -> Option A

5. Quick Check:

   Lazy removal with hash map + heaps = efficient median update [OK]

Hint: Lazy removal with hash map keeps heaps efficient [OK]

Common Mistakes:

• Trying direct removal from heaps
• Using arrays causing slow operations
• Ignoring complexity of balanced trees