Kth Largest Element Using Max Heap in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to build a max heap from the list.

DSA Python

import heapq

nums = [3, 2, 1, 5, 6, 4]
max_heap = [-n for n in nums]
heapq.[1](max_heap)
print(max_heap)

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

Aheapify

Bheappush

Cheappop

Dheapreplace

Attempts:

3 left

💡 Hint

Common Mistakes

Using heappush instead of heapify to build the heap.

Not negating numbers to simulate max heap.

The heapify function converts a list into a heap in-place. Here, we negate numbers to simulate a max heap using Python's min heap.

2fill in blank

medium

Complete the code to pop the largest element from the max heap.

DSA Python

largest = -heapq.[1](max_heap)
print(largest)

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

Aheappush

Bheapreplace

Cheapify

Dheappop

Attempts:

3 left

💡 Hint

Common Mistakes

Using heapify instead of heappop to remove element.

Not negating the popped value to get the original number.

heappop removes and returns the smallest element from the heap. Since we negated numbers, it returns the largest original number when negated back.

3fill in blank

hard

Fix the error in the code to find the kth largest element using max heap.

DSA Python

def find_kth_largest(nums, k):
    max_heap = [-n for n in nums]
    heapq.heapify(max_heap)
    for _ in range(k-1):
        heapq.[1](max_heap)
    return -max_heap[0]

print(find_kth_largest([3,2,3,1,2,4,5,5,6], 4))

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

Aheappop

Bheappush

Cheapreplace

Dheapify

Attempts:

3 left

💡 Hint

Common Mistakes

Using heappush instead of heappop in the loop.

Using heapify inside the loop causing errors.

To remove the top elements k-1 times, we use heappop. This pops the largest elements (negated) from the heap.

4fill in blank

hard

Fill both blanks to create a max heap and pop the largest element.

DSA Python

nums = [10, 20, 15]
max_heap = [[1] for n in nums]
heapq.[2](max_heap)
largest = -heapq.heappop(max_heap)
print(largest)

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

A-n

Bn

Cheapify

Dheappush

Attempts:

3 left

💡 Hint

Common Mistakes

Not negating numbers before heapify.

Using heappush instead of heapify to build heap.

Negate numbers to simulate max heap and use heapify to build the heap.

5fill in blank

hard

Fill all three blanks to find the kth largest element using max heap.

DSA Python

def find_kth_largest(nums, k):
    max_heap = [[1] for n in nums]
    heapq.[2](max_heap)
    for _ in range(k-1):
        heapq.[3](max_heap)
    return -max_heap[0]

print(find_kth_largest([7, 10, 4, 3, 20, 15], 3))

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

A-n

Bheapify

Cheappop

Dheappush

Attempts:

3 left

💡 Hint

Common Mistakes

Not negating numbers before heapify.

Using heappush instead of heappop in the loop.

Using heapify inside the loop.

Negate numbers to simulate max heap, build heap with heapify, then pop k-1 times with heappop.

Practice

1. What is the main idea behind finding the kth largest element using a max heap?

easy

A. Build a max heap from the array and extract the max element k times

B. Build a min heap and extract the min element k times

C. Sort the array and pick the kth element from the start

D. Use a queue to store elements and dequeue k times

Solution

1. Step 1: Understand max heap property

   A max heap always keeps the largest element at the root, so extracting the root gives the largest element.

2. Step 2: Extract max k times to find kth largest

   By removing the root k times, the last extracted element is the kth largest.

3. Final Answer:

   Build a max heap and extract max k times -> Option A

4. Quick Check:

   Max heap extraction = kth largest element [OK]

Hint: Extract max from max heap k times to get kth largest [OK]

Common Mistakes:

• Using min heap instead of max heap
• Sorting and picking wrong index
• Extracting fewer or more than k times

2. Which of the following Python code snippets correctly builds a max heap from a list arr?

easy

A. import heapq
max_heap = [-x for x in arr]
heapq.heapify(max_heap)

B. import heapq
heapq.heapify(arr)

C. arr.sort(reverse=True)

D. max_heap = arr
max_heap.sort()

Solution

1. Step 1: Recognize Python's heapq is a min heap

   Python's heapq creates a min heap by default, so to simulate max heap, we invert values.

2. Step 2: Invert values and heapify

   By negating each element and heapifying, we get a max heap behavior.

3. Final Answer:

   Negate elements and heapify using heapq -> Option A

4. Quick Check:

   Negate + heapify = max heap in Python [OK]

Hint: Negate values before heapify to simulate max heap in Python [OK]

Common Mistakes:

• Using heapq.heapify directly without negation
• Sorting instead of heapifying
• Assuming heapq builds max heap by default

3. What is the output of the following code snippet?

import heapq
arr = [3, 1, 5, 12, 2, 11]
max_heap = [-x for x in arr]
heapq.heapify(max_heap)
k = 3
for _ in range(k - 1):
    heapq.heappop(max_heap)
kth_largest = -heapq.heappop(max_heap)
print(kth_largest)

medium

A. 3

B. 11

C. 5

D. 12

Solution

1. Step 1: Build max heap by negating and heapifying

   Negated list: [-3, -1, -5, -12, -2, -11], heapified to max heap behavior.

2. Step 2: Pop max element k-1=2 times

   First pop: removes -12 (12), second pop: removes -11 (11).

3. Step 3: Pop again to get kth largest

   Third pop removes -5 (5), which is the 3rd largest element.

4. Final Answer:

   5 -> Option C

5. Quick Check:

   3rd largest in [3,1,5,12,2,11] = 5 [OK]

Hint: Pop max heap k times, last pop is kth largest [OK]

Common Mistakes:

• Not negating values before heapify
• Popping fewer or more times
• Confusing min heap with max heap

4. Identify the error in this code snippet to find the 2nd largest element using a max heap:

import heapq
arr = [4, 7, 6, 3]
heapq.heapify(arr)
k = 2
for _ in range(k):
    largest = heapq.heappop(arr)
print(largest)

medium

A. heapq.heapify creates a min heap, not max heap

B. All of the above

C. Missing negation of elements to simulate max heap

D. Loop should run k-1 times, not k

Solution

1. Step 1: Recognize heapq creates min heap by default

   Without negation, heapq.heapify creates a min heap, so popping gives smallest elements.

2. Step 2: Loop runs k times instead of k-1

   To get the kth largest, pop k-1 times, then pop once more for answer.

3. Step 3: Negation missing to simulate max heap

   Negating elements before heapify is needed to simulate max heap behavior.

4. Final Answer:

   All of the above -> Option B

5. Quick Check:

   Min heap + wrong loop + no negation = error [OK]

Hint: Negate values and pop k-1 times before final pop [OK]

Common Mistakes:

• Not negating values for max heap
• Looping wrong number of times
• Assuming heapq builds max heap by default

5. Given a large list arr with 1 million elements, which approach is most efficient to find the 10th largest element using a max heap?

hard

A. Build a max heap of all elements and pop 10 times

B. Use a max heap but pop all elements until the 10th largest

C. Sort the entire list and pick the 10th largest

D. Build a min heap of size 10 and iterate through elements

Solution

1. Step 1: Understand max heap approach cost

   Building max heap of 1 million and popping 10 times is costly (O(n) + O(k log n)).

2. Step 2: Use min heap of size 10 for efficiency

   Maintain a min heap of size 10 while iterating; this keeps track of top 10 largest efficiently (O(n log k)).

3. Step 3: Compare options

   Sorting entire list is O(n log n), more expensive. Popping all elements is unnecessary.

4. Final Answer:

   Build a min heap of size 10 and iterate -> Option D

5. Quick Check:

   Min heap size k = efficient kth largest [OK]

Hint: Use min heap of size k for large data to find kth largest fast [OK]

Common Mistakes:

• Building max heap of entire large list
• Sorting entire list unnecessarily
• Popping all elements instead of just k times