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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the max heap for the lower half of numbers.

DSA Go

maxHeap := &IntHeap[1]

Drag options to blanks, or click blank then click option'

A[]int{}

B{}

Cmake([]int, 0)

Dnil

Attempts:

3 left

💡 Hint

Common Mistakes

Using nil instead of an empty slice causes runtime errors when pushing.

Using {} is invalid syntax for slice initialization.

In Go, to initialize an empty slice for the heap, use []int{}.

2fill in blank

medium

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

DSA Go

heap.[1](maxHeap, num)

Drag options to blanks, or click blank then click option'

AInsert

Bpush

CAdd

DPush

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase 'push' causes compilation errors.

Using 'Add' or 'Insert' are not valid heap package methods.

The heap package in Go uses 'Push' with uppercase P to add elements.

3fill in blank

hard

Fix the error in the code to pop the smallest number from the min heap.

DSA Go

smallest := heap.[1](minHeap).(int)

Drag options to blanks, or click blank then click option'

Apop

BRemove

CPop

DExtract

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase 'pop' causes compilation errors.

Using 'Remove' or 'Extract' are not valid heap package methods.

The heap package uses 'Pop' with uppercase P to remove the smallest element.

4fill in blank

hard

Fill both blanks to balance the heaps by moving the top element from maxHeap to minHeap.

DSA Go

top := heap.[1](maxHeap).(int)
heap.[2](minHeap, top)

Drag options to blanks, or click blank then click option'

APop

BPush

CRemove

DInsert

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Remove' or 'Insert' instead of 'Pop' and 'Push'.

Mixing up the order of operations.

Use 'Pop' to remove from maxHeap and 'Push' to add to minHeap.

5fill in blank

hard

Fill all three blanks to calculate the median from the two heaps.

DSA Go

if maxHeap.Len() > minHeap.Len() {
    median := float64(maxHeap.[1]().(int))
    return median
} else {
    median := (float64(maxHeap.[2]().(int)) + float64(minHeap.[3]().(int))) / 2
    return median
}

Drag options to blanks, or click blank then click option'

APeek

BPop

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'Pop' removes elements and changes the heaps incorrectly.

Not accessing the top elements properly.

Use 'Peek' (a custom method) to look at the top element without removing it for median calculation.

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