Median of Data Stream Using Two Heaps in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind using two heaps to find the median of a data stream?

Use a max-heap to store the smaller half of numbers and a min-heap to store the larger half. This way, the median can be quickly found from the top elements of the heaps.

beginner

Why do we balance the sizes of the two heaps when finding the median?

Balancing ensures that the heaps differ in size by at most one, so the median is either the top of one heap or the average of the tops of both heaps.

beginner

In the two heaps approach, which heap stores the smaller half of the numbers?

The max-heap stores the smaller half of the numbers, so its top is the largest of the smaller half.

intermediate

How do you find the median when the total number of elements is even using two heaps?

The median is the average of the top elements of the max-heap and min-heap.

intermediate

What is the time complexity of adding a number and finding the median using two heaps?

Adding a number takes O(log n) due to heap insertion, and finding the median is O(1) because it uses the top elements of the heaps.

Which heap stores the larger half of the numbers in the two heaps approach?

A Min-heap

B Max-heap

C Both heaps store the same numbers

D None of the above

The min-heap stores the larger half of the numbers, so its top is the smallest of the larger half.

What do you do if the max-heap has two more elements than the min-heap after insertion?

A Do nothing

B Remove the top from min-heap and add it to max-heap

C Remove the top from max-heap and add it to min-heap

D Clear both heaps

To balance the heaps, move the top element from the larger max-heap to the min-heap.

If the total number of elements is odd, where does the median come from?

A Top of min-heap

B Top of max-heap

C Average of tops of both heaps

D Bottom of max-heap

When odd, the max-heap has one extra element, so the median is its top.

What is the main advantage of using two heaps for median of data stream?

A Uses less memory than arrays

B Only works for sorted data

C No need to balance heaps

D Fast insertion and fast median retrieval

Two heaps allow O(log n) insertion and O(1) median retrieval.

Which operation is NOT needed when adding a new number to the two heaps?

A Sort the entire data stream

B Insert into one heap

C Balance the heaps

D Compare with max-heap top

Sorting the entire data stream is not needed; heaps keep order efficiently.

Explain how two heaps are used to find the median in a data stream.

Think about splitting numbers into two groups and how to keep track of middle values.

Describe the steps to add a new number to the two heaps and maintain the median.

Focus on where to put the new number and how to keep heaps balanced.

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