Min Heap vs Max Heap When to Use Which in DSA Go - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a min heap by choosing the correct comparison operator.

DSA Go

func (h MinHeap) Less(i, j int) bool {
    return h[i] [1] h[j]
}

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

A>

B!=

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < causes a max heap behavior.

Using == or != does not define order.

In a min heap, the parent node is smaller than its children, so we use the less than operator < to compare.

2fill in blank

medium

Complete the code to create a max heap by choosing the correct comparison operator.

DSA Go

func (h MaxHeap) Less(i, j int) bool {
    return h[i] [1] h[j]
}

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

A>

B==

C<

D!=

Attempts:

3 left

💡 Hint

Common Mistakes

Using < instead of > causes a min heap behavior.

Using == or != does not define order.

In a max heap, the parent node is greater than its children, so we use the greater than operator > to compare.

3fill in blank

hard

Fix the error in the code to pop the root element from a min heap.

DSA Go

func (h *MinHeap) Pop() interface{} {
    old := *h
    n := len(old)
    x := old[[1]]
    *h = old[0 : n-1]
    return x
}

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

An-1

B0

Cn

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n or n-1 causes out of range or wrong element removal.

Using 1 removes the second element, not the root.

The root element of a heap is always at index 0, so we pop old[0].

4fill in blank

hard

Fill both blanks to build a min heap from a slice of integers.

DSA Go

h := MinHeap[1]
heap.Init(&h)
return h[2]

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

A{data}

Bdata

C&h

D*h

Attempts:

3 left

💡 Hint

Common Mistakes

Using &h instead of data causes type mismatch.

Returning &h instead of h returns a pointer, not the struct.

We assign the slice 'data' to the heap struct and return the heap struct with the data inside curly braces.

5fill in blank

hard

Fill all three blanks to push a new value into a max heap and maintain heap property.

DSA Go

func (h *MaxHeap) Push(x interface{}) {
    *h = append(*h, [1])
    heap.[2](h)
}

func (h *MaxHeap) Pop() interface{} {
    old := *h
    n := len(old)
    x := old[[3]]
    *h = old[0 : n-1]
    return x
}

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

Ax

BFix

C0

DPush

Attempts:

3 left

💡 Hint

Common Mistakes

Appending wrong variable or calling heap.Fix instead of heap.Push.

Popping element at wrong index.

We append the new value x, then call heap.Push to maintain the heap property, and pop the root at index 0.

Practice

1. What is the main difference between a Min Heap and a Max Heap?

easy

A. Min Heap always has the smallest element at the root, Max Heap has the largest at the root

B. Min Heap stores only positive numbers, Max Heap stores only negative numbers

C. Min Heap is a sorted array, Max Heap is a linked list

D. Min Heap and Max Heap are the same data structure

Solution

1. Step 1: Understand heap root property

   Min Heap keeps the smallest element at the root node, so it can be accessed quickly. Max Heap keeps the largest element at the root.

2. Step 2: Compare Min Heap and Max Heap behavior

   This difference defines their use cases: Min Heap for quick access to minimum, Max Heap for maximum.

3. Final Answer:

   Min Heap always has the smallest element at the root, Max Heap has the largest at the root -> Option A

4. Quick Check:

   Root element defines heap type = A [OK]

Hint: Root element smallest? Min Heap; largest? Max Heap [OK]

Common Mistakes:

• Confusing Min Heap with sorted array
• Thinking heaps store only positive or negative numbers
• Believing Min Heap and Max Heap are identical

2. Which of the following is the correct way to declare a Min Heap in Go using container/heap package?

easy

A. var h = make([]int, 0) // slice without heap.Interface

B. var h = &IntHeap{} // IntHeap implements heap.Interface with Less for min heap

C. var h = &IntHeap{} // IntHeap implements heap.Interface with Less for max heap

D. var h = new(MaxHeap) // MaxHeap is built-in type

Solution

1. Step 1: Understand Go heap.Interface usage

   In Go, to create a Min Heap, you implement heap.Interface with Less method returning true if element i is less than j.

2. Step 2: Identify correct declaration

   var h = &IntHeap{} // IntHeap implements heap.Interface with Less for min heap shows a pointer to IntHeap which implements heap.Interface correctly for Min Heap. Others are incorrect or incomplete.

3. Final Answer:

   var h = &IntHeap{} // IntHeap implements heap.Interface with Less for min heap -> Option B

4. Quick Check:

   heap.Interface with Less for min heap = C [OK]

Hint: Min Heap needs Less(i,j) true if i

Common Mistakes:

• Using slice without implementing heap.Interface
• Confusing Less method for max heap
• Assuming MaxHeap is built-in type

3. What will be the output of the following Go code snippet that uses a Max Heap?

package main

import (
  "container/heap"
  "fmt"
)

type IntHeap []int

func (h IntHeap) Len() int           { return len(h) }
func (h IntHeap) Less(i, j int) bool { return h[i] > h[j] } // Max Heap
func (h IntHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }

func (h *IntHeap) Push(x interface{}) {
  *h = append(*h, x.(int))
}

func (h *IntHeap) Pop() interface{} {
  old := *h
  n := len(old)
  x := old[n-1]
  *h = old[0 : n-1]
  return x
}

func main() {
  h := &IntHeap{2, 5, 1}
  heap.Init(h)
  heap.Push(h, 10)
  fmt.Println(heap.Pop(h))
}

medium

A. 10

B. 5

C. 2

D. 1

Solution

1. Step 1: Understand Max Heap property in Less method

   The Less method returns true if h[i] > h[j], so the largest element is at the root.

2. Step 2: Trace heap operations

   Initial heap: [2,5,1]. After heap.Init, heap property is established. Push 10 adds 10. Pop removes and returns the largest element, which is 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Max Heap root is largest = 10 [OK]

Hint: Max Heap pops largest element first [OK]

Common Mistakes:

• Confusing Min Heap and Max Heap Less method
• Assuming Pop returns smallest element
• Not initializing heap before Push

4. The following Go code tries to implement a Min Heap but does not work correctly. What is the bug?

type IntHeap []int

func (h IntHeap) Len() int           { return len(h) }
func (h IntHeap) Less(i, j int) bool { return h[i] > h[j] } // Intended Min Heap
func (h IntHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }

func (h *IntHeap) Push(x interface{}) {
  *h = append(*h, x.(int))
}

func (h *IntHeap) Pop() interface{} {
  old := *h
  n := len(old)
  x := old[n-1]
  *h = old[0 : n-1]
  return x
}

medium

A. Swap method swaps elements incorrectly

B. Push method does not append correctly

C. Less method uses '>' instead of '<', so it creates a Max Heap

D. Pop method returns wrong element

Solution

1. Step 1: Analyze Less method for heap type

   The Less method returns true if h[i] > h[j], which means it treats larger elements as higher priority, creating a Max Heap, not Min Heap.

2. Step 2: Confirm other methods are correct

   Push appends correctly, Pop returns last element, Swap swaps correctly. The bug is only in Less method.

3. Final Answer:

   Less method uses '>' instead of '<', so it creates a Max Heap -> Option C

4. Quick Check:

   Less defines heap type; '>' means Max Heap [OK]

Hint: Less method '<' for Min Heap, '>' for Max Heap [OK]

Common Mistakes:

• Mixing up Less method logic
• Blaming Push or Pop methods incorrectly
• Ignoring heap.Interface contract

5. You want to efficiently track the top 3 highest scores from a stream of player scores in Go. Which heap type should you use and why?

hard

A. Use a Max Heap of unlimited size to store all scores

B. Use a Max Heap of size 3 to keep the top scores by removing larger ones

C. Use a Min Heap of unlimited size to store all scores

D. Use a Min Heap of size 3 to keep the top scores by removing smaller ones

Solution

1. Step 1: Understand the problem of tracking top 3 highest scores

   We want to keep only the top 3 highest scores efficiently as new scores arrive.

2. Step 2: Choose heap type and size

   A Min Heap of size 3 keeps the smallest of the top scores at the root. When a new score is higher than the root, replace root and heapify. This keeps top 3 highest scores efficiently.

3. Final Answer:

   Use a Min Heap of size 3 to keep the top scores by removing smaller ones -> Option D

4. Quick Check:

   Min Heap size 3 tracks top scores by removing smaller [OK]

Hint: Min Heap size k tracks top k largest by removing smaller [OK]

Common Mistakes:

• Using Max Heap and removing larger scores incorrectly
• Using unlimited size heaps wasting memory
• Not limiting heap size to k