Median of Data Stream Using Two Heaps in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the median after inserting these numbers?

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

DSA C

Insert 5
Insert 15
Insert 1
Insert 3
Find median

A 4.0

B 3.0

C 5.0

D 6.0

💡 Hint

Remember to balance the two heaps after each insertion and calculate median accordingly.

After inserting 5, 15, 1, 3, the two heaps balance to hold the lower half and upper half of numbers. The median is the average of the two middle numbers: (3 + 5) / 2 = 4.0.

🧠 Conceptual

intermediate

Why use two heaps to find median in a data stream?

What is the main reason for using two heaps (a max heap and a min heap) to find the median of a data stream efficiently?

A To keep track of the lower half and upper half of numbers separately for quick median calculation

B To sort all numbers in ascending order continuously

C To store all numbers twice for redundancy

D To use less memory by discarding half the numbers

💡 Hint

Think about how median relates to the middle values in a sorted list.

The two heaps split the data stream into lower and upper halves. The max heap stores the smaller half, and the min heap stores the larger half. This allows quick access to the middle values needed for median calculation.

🔧 Debug

advanced

Identify the error in this median update code snippet

What error will this code cause when updating the median after inserting a new number into the two heaps? ```c if (maxHeapSize > minHeapSize + 1) { int val = extractMax(maxHeap); insertMin(minHeap, val); } else if (minHeapSize > maxHeapSize) { int val = extractMin(minHeap); insertMax(maxHeap, val); } median = (maxHeapSize == minHeapSize) ? (maxHeap[0] + minHeap[0]) / 2 : maxHeap[0]; ``` Assume maxHeap and minHeap are arrays representing heaps with root at index 0.

DSA C

if (maxHeapSize > minHeapSize + 1) {
    int val = extractMax(maxHeap);
    insertMin(minHeap, val);
} else if (minHeapSize > maxHeapSize) {
    int val = extractMin(minHeap);
    insertMax(maxHeap, val);
}
median = (maxHeapSize == minHeapSize) ? (maxHeap[0] + minHeap[0]) / 2 : maxHeap[0];

A Integer division causes incorrect median when sum is odd

B Median calculation uses wrong heap roots

C Extracting max from maxHeap is invalid operation

D Heap sizes are not updated after extraction and insertion

💡 Hint

Check if heap size variables reflect the actual number of elements after moving values.

The code moves elements between heaps but does not update the heap size variables. This causes incorrect median calculation and possible out-of-bounds errors.

❓ Predict Output

advanced

What is the median after this sequence of insertions?

Insert the numbers 2, 4, 7, 1, 5, 3 into the data stream using two heaps. What is the median after all insertions?

DSA C

Insert 2
Insert 4
Insert 7
Insert 1
Insert 5
Insert 3
Find median

A 4.5

B 3.5

C 3.0

D 4.0

💡 Hint

Balance heaps after each insertion and find the middle two numbers for even count.

After inserting all numbers, the two heaps split the data into halves: lower half [1,2,3] and upper half [4,5,7]. Median is average of 3 and 4 = 3.5.

🚀 Application

expert

How to efficiently find median in a large real-time data stream?

You receive a large stream of numbers in real-time and need to find the median after each new number arrives. Which approach is the most efficient and scalable?

A Use a balanced binary search tree to insert numbers and find median by in-order traversal

B Store all numbers in an array and sort it after each insertion to find the median

C Use two heaps (max heap and min heap) to keep track of lower and upper halves and balance them after each insertion

D Keep only the last 100 numbers and calculate median from them ignoring older data

💡 Hint

Consider time complexity of insertion and median retrieval for large data streams.

Two heaps allow O(log n) insertion and O(1) median retrieval, making them efficient for large real-time data streams. Sorting after each insertion or traversing trees is slower. Ignoring data loses accuracy.

Practice

1. What is the main purpose of using two heaps (a max-heap and a min-heap) to find 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 remove duplicate numbers from the stream

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

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 D

4. Quick Check:

   Two heaps split data halves = To efficiently keep track of the lower half and upper half of the numbers separately [OK]

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

Common Mistakes:

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

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

easy

A. Insert into the heap based on comparison with max-heap top

B. Insert into min-heap only

C. Insert into max-heap only

D. Always insert into min-heap first, then balance heaps

Solution

1. Step 1: Decide where to insert the new number

   If the new number is less than or equal to the max-heap top, insert into max-heap; otherwise, insert into min-heap.

2. Step 2: Balance the heaps if needed

   After insertion, balance the heaps so their sizes differ by at most one.

3. Final Answer:

   Insert into the heap based on comparison with max-heap top -> Option A

4. Quick Check:

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

Hint: Compare new number with max-heap top to choose heap [OK]

Common Mistakes:

• Always inserting into one heap without condition
• Not balancing heaps after insertion
• Inserting into wrong heap based on min-heap top

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

Insert 5
Insert 15
Insert 1
Insert 3
Find median

medium

A. 5

B. 3

C. 4

D. 6

Solution

1. Step 1: Insert numbers and balance heaps

   Insert 5 into max-heap ([5]); insert 15 into min-heap ([15]); insert 1 (<=5) into max-heap ([5,1] top=5); insert 3 (<=5) into max-heap ([5,3,1] top=5), sizes 3 vs 1, balance: min.push(5) ([15,5] top=5), max.pop() ([3,1] top=3).

2. Step 2: Calculate median from heaps

   Heaps sizes equal (2 and 2), median = average of max-heap top (3) and min-heap top (5) = (3+5)/2 = 4.

3. Final Answer:

   4 -> Option C

4. Quick Check:

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

Hint: Median is average when heaps equal size [OK]

Common Mistakes:

• Taking median as max-heap top only
• Not balancing heaps after insertions
• Calculating median incorrectly as 5 or 3

4. The following code snippet tries to maintain median using two heaps but has a bug. What is the bug?

void addNumber(int num, priority_queue<int>& maxHeap, priority_queue& minHeap) {
    if (maxHeap.empty() || num <= maxHeap.top()) {
        maxHeap.push(num);
    } else {
        minHeap.push(num);
    }
    if (maxHeap.size() > minHeap.size() + 1) {
        minHeap.push(maxHeap.top());
        maxHeap.pop();
    } else if (minHeap.size() > maxHeap.size() + 1) {
        maxHeap.push(minHeap.top());
        minHeap.pop();
    }
}

medium

A. The balancing pops top element before pushing, causing wrong order

B. The condition to insert into maxHeap is incorrect

C. The heaps are not balanced after insertion

D. The function does not return the median

Solution

1. Step 1: Analyze balancing logic

   When balancing, the code pops the top element of one heap before pushing the new top to the other, which moves the wrong element.

2. Step 2: Explain correct balancing

   Correct approach is to push heap.top() first, then pop() to transfer the correct top element.

3. Final Answer:

   The balancing pops top element before pushing, causing wrong order -> Option A

4. Quick Check:

   Push top before pop to balance heaps correctly [OK]

Hint: Push top before pop when balancing heaps [OK]

Common Mistakes:

• Popping before pushing the correct top
• Ignoring heap size difference conditions
• Not handling empty heaps properly

5. You have a data stream of integers and want to find the median after each insertion. Using two heaps, how would you handle the case when the total number of elements is odd?

hard

A. The median is always the top of the min-heap

B. The median is the top element of the heap with one extra element

C. The median is always the top of the max-heap

D. The median is the average of the tops of both heaps

Solution

1. Step 1: Understand heap size difference

   When total elements are odd, one heap will have one more element than the other.

2. Step 2: Determine median from heaps

   The median is the top element of the heap that has the extra element, because it holds the middle value.

3. Final Answer:

   The median is the top element of the heap with one extra element -> Option B

4. Quick Check:

   Odd count median = top of larger heap [OK]

Hint: Median is top of bigger heap when odd count [OK]

Common Mistakes:

• Averaging tops when count is odd
• Assuming max-heap always has extra element
• Ignoring heap size difference