Why Median of Data Stream Using Two Heaps in DSA Go?

The Big Idea

What if you could find the middle value instantly, no matter how many numbers you add?

The Scenario

Imagine you are tracking the middle value of a growing list of numbers, like recording daily temperatures. You write down every number and then sort the entire list each time you want to find the middle value.

The Problem

Sorting the whole list every time is slow and tiresome, especially as the list grows longer. It wastes time and can cause mistakes when done manually or with simple code.

The Solution

Using two heaps (one for smaller half, one for larger half) keeps the numbers balanced and lets you quickly find the middle value without sorting everything again and again.

Before vs After

✗ Before

numbers = []
numbers.append(newNumber)
numbers.sort()
median = numbers[len(numbers)//2]

✓ After

smallHeap.Push(newNumber)
// Balance heaps
median = smallHeap.Top()

What It Enables

This method lets you find the median instantly as new numbers come in, making real-time data analysis smooth and fast.

Real Life Example

Streaming apps use this to show live median ratings or prices without delays, even when millions of users add data continuously.

Key Takeaways

Manual sorting each time is slow and inefficient.

Two heaps keep data balanced for quick median access.

Enables fast, real-time median calculation on growing data.

Practice

1. What is the main purpose of using two heaps (a max-heap and a min-heap) in finding 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 efficiently keep track of the lower half and upper half of the numbers separately

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 C

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 one heap only
• Confusing max-heap and min-heap roles

2. Which of the following Go code snippets correctly initializes a max-heap and a min-heap for integers using container/heap package?

easy

A. maxHeap := &IntMaxHeap{}; minHeap := &IntMinHeap{}

B. maxHeap := heap.New(IntMaxHeap{}); minHeap := heap.New(IntMinHeap{})

C. maxHeap := &IntMinHeap{}; minHeap := &IntMaxHeap{}

D. maxHeap := make([]int, 0); minHeap := make([]int, 0)

Solution

1. Step 1: Recognize heap initialization in Go

   Heaps are initialized as pointers to custom heap types that implement heap.Interface.

2. Step 2: Identify correct syntax

   Using &IntMaxHeap{} and &IntMinHeap{} creates pointers to heap structs ready for heap operations.

3. Final Answer:

   maxHeap := &IntMaxHeap{}; minHeap := &IntMinHeap{} -> Option A

4. Quick Check:

   Pointer to heap struct = A [OK]

Hint: Use pointers to heap structs for container/heap [OK]

Common Mistakes:

• Using heap.New which does not exist
• Swapping max-heap and min-heap types
• Initializing as slices without heap.Interface

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. 5

B. 8

C. 3

D. 4

Solution

1. Step 1: Insert numbers and balance heaps

   Insert 5 (maxHeap top=5); insert 15 (minHeap top=15); insert 1 ≤5 (maxHeap top=5, contains 5,1); insert 3 ≤5 (maxHeap top=5, contains 5,1,3), balance: len=3 >1+1, pop 5 -> push minHeap top=5 (15,5), maxHeap top=3 (1,3).

2. Step 2: Calculate median from heaps

   Heaps sizes equal (2 each). Median = average of maxHeap top (3) and minHeap top (5) = (3+5)/2 = 4.

3. Final Answer:

   4 -> Option D

4. Quick Check:

   Median = (3+5)/2 = 4 [OK]

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

Common Mistakes:

• Taking median as maxHeap top only
• Not balancing heaps after insertion
• Calculating median as sum instead of average

4. In the following Go code snippet for median calculation, what is the bug?

func addNum(num int) {
  if maxHeap.Len() == 0 || num <= (*maxHeap)[0] {
    heap.Push(maxHeap, num)
  } else {
    heap.Push(minHeap, num)
  }
  if maxHeap.Len() > minHeap.Len()+1 {
    heap.Push(minHeap, heap.Pop(maxHeap))
  } else if minHeap.Len() > maxHeap.Len() {
    heap.Push(maxHeap, heap.Pop(minHeap))
  }
}

medium

A. Heaps are not balanced after insertion

B. Comparison operator should be < instead of <=

C. heap.Pop returns interface{}, needs type assertion before Push

D. maxHeap and minHeap are swapped in conditions

Solution

1. Step 1: Analyze balancing logic

   The first condition correctly balances when maxHeap is too large (> minHeap.Len()+1). But the second condition checks minHeap.Len() > maxHeap.Len(), which triggers rebalance correctly when minHeap is larger.

2. Step 2: Why heaps are not balanced

   Original code used > maxHeap.Len() + 1 which fails to rebalance when minHeap is exactly one larger. Changing to > maxHeap.Len() fixes this.

3. Final Answer:

   Heaps are not balanced after insertion -> Option A

4. Quick Check:

   minHeap balance: > max.Len() [OK]

Hint: minHeap balance when len(min) > len(max) [OK]

Common Mistakes:

• Using +1 symmetrically for both balance conditions
• Wrong size difference in minHeap check
• Not balancing after every insertion

5. You want to find the median of a large data stream where numbers can be negative, zero, or positive. Which approach using two heaps ensures correct median calculation even with duplicates and negative values?

hard

A. Use a single max-heap and pop elements until median is found

B. Use a max-heap for lower half and min-heap for upper half, balance sizes, and calculate median from tops

C. Sort all numbers in an array and pick middle element(s) each time

D. Use two min-heaps and merge their outputs to find median

Solution

1. Step 1: Choose correct heap types for halves

   Max-heap stores lower half (including negatives), min-heap stores upper half, handling all values correctly.

2. Step 2: Balance heaps and compute median

   Keep heaps balanced in size; median is top of max-heap if odd count, or average of tops if even count.

3. Final Answer:

   Use a max-heap for lower half and min-heap for upper half, balance sizes, and calculate median from tops -> Option B

4. Quick Check:

   Two heaps balanced handle negatives and duplicates [OK]

Hint: Balance max and min heaps for all number ranges [OK]

Common Mistakes:

• Using two min-heaps which can't split halves properly
• Sorting entire stream each time (inefficient)
• Using single heap only