Median of Data Stream Using Two Heaps in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a number into the max heap.

DSA C

void insertMaxHeap(int heap[], int *size, int value) {
    heap[++(*size)] = [1];
    int i = *size;
    while (i > 1 && heap[i / 2] < heap[i]) {
        int temp = heap[i];
        heap[i] = heap[i / 2];
        heap[i / 2] = temp;
        i /= 2;
    }
}

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Ai

Bsize

Cheap[*size]

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong variable instead of 'value' to insert.

Not incrementing the size before insertion.

We insert the new value at the end of the heap array before heapifying up.

2fill in blank

medium

Complete the code to balance the sizes of the two heaps.

DSA C

void balanceHeaps(int maxHeap[], int *maxSize, int minHeap[], int *minSize) {
    if (*maxSize > *minSize + 1) {
        int value = extractMax(maxHeap, maxSize);
        insertMinHeap(minHeap, minSize, [1]);
    } else if (*minSize > *maxSize) {
        int value = extractMin(minHeap, minSize);
        insertMaxHeap(maxHeap, maxSize, value);
    }
}

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A*minSize

B*maxSize

Cvalue

DmaxHeap[1]

Attempts:

3 left

💡 Hint

Common Mistakes

Inserting the wrong variable into minHeap.

Using heap size variables instead of the extracted value.

The extracted max value from maxHeap should be inserted into minHeap to balance sizes.

3fill in blank

hard

Fix the error in the median calculation code.

DSA C

double getMedian(int maxHeap[], int maxSize, int minHeap[], int minSize) {
    if (maxSize == minSize) {
        return (maxHeap[1] + [1]) / 2.0;
    } else {
        return maxHeap[1];
    }
}

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AminHeap[1]

BmaxHeap[0]

CminHeap[0]

DmaxHeap[maxSize]

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 0 for heap root which is incorrect in 1-based heap arrays.

Using maxHeap elements instead of minHeap for the second median value.

Median is average of max of maxHeap and min of minHeap, which are at index 1 in heaps.

4fill in blank

hard

Fill both blanks to correctly insert a new number into the appropriate heap.

DSA C

void addNumber(int num, int maxHeap[], int *maxSize, int minHeap[], int *minSize) {
    if (*maxSize == 0 || num <= maxHeap[1]) {
        insertMaxHeap(maxHeap, maxSize, [1]);
    } else {
        insertMinHeap(minHeap, minSize, [2]);
    }
    balanceHeaps(maxHeap, maxSize, minHeap, minSize);
}

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Anum

B*maxSize

C*minSize

DmaxHeap[1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using heap sizes instead of the number to insert.

Inserting wrong heap root values instead of 'num'.

The new number 'num' is inserted into maxHeap or minHeap based on comparison.

5fill in blank

hard

Fill both blanks to correctly extract the max from maxHeap.

DSA C

int extractMax(int heap[], int *size) {
    int max = heap[1];
    heap[1] = heap[[1]];
    (*size)--;
    int i = 1;
    while (i * 2 <= *size) {
        int left = i * 2;
        int right = i * 2 + 1;
        int largest = i;
        if (heap[left] > heap[largest]) largest = left;
        if (right <= *size && heap[right] > heap[largest]) largest = right;
        if (largest == i) break;
        int temp = heap[i];
        heap[i] = heap[largest];
        heap[largest] = temp;
        i = largest;
    }
    return [2];
}

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A1

Bmax

C*size

Dheap[1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong index to replace root.

Returning wrong variable instead of max.

Replace root with last element at index *size, decrease size, then return max.

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