Top K Frequent Elements Using Heap in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to count the frequency of each element in the list.

DSA Python

from collections import Counter

def top_k_frequent(nums, k):
    freq = Counter(nums)
    return freq[1]

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

A.keys()

B.items()

C.values()

D.most_common()

Attempts:

3 left

💡 Hint

Common Mistakes

Using .keys() returns only elements without counts.

Using .values() returns only counts without elements.

Using .most_common() returns a list, not a dict view.

We use freq.items() to get pairs of (element, frequency) for processing.

2fill in blank

medium

Complete the code to push frequency and element pairs into the heap.

DSA Python

import heapq

heap = []
for num, count in freq.items():
    heapq.heappush(heap, ([1], num))

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

A-count

Bnum

C-num

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Pushing count directly creates a min heap, not max heap.

Pushing num instead of count does not sort by frequency.

We push (-count, num) to create a max heap by frequency using Python's min heap.

3fill in blank

hard

Fix the error in extracting the top k frequent elements from the heap.

DSA Python

result = []
for _ in range(k):
    freq, num = heapq.heappop(heap)
    result.append([1])
return result

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

Afreq

B-freq

Cnum

D-num

Attempts:

3 left

💡 Hint

Common Mistakes

Appending freq adds negative counts, not elements.

Appending -freq would give positive counts, not elements.

We append num because it is the element, freq is the negative count.

4fill in blank

hard

Fill both blanks to complete the function that returns the top k frequent elements using a heap.

DSA Python

def top_k_frequent(nums, k):
    freq = Counter(nums)
    heap = []
    for num, count in freq.items():
        heapq.heappush(heap, ([1], num))
    result = []
    for _ in range(k):
        freq, num = heapq.heappop(heap)
        result.append([2])
    return result

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

A-count

Bnum

Ccount

D-num

Attempts:

3 left

💡 Hint

Common Mistakes

Using count instead of -count breaks max heap logic.

Appending freq instead of num adds counts, not elements.

Push negative count for max heap and append num to result.

5fill in blank

hard

Fill all three blanks to create a complete function that returns the top k frequent elements using a heap.

DSA Python

import heapq
from collections import Counter

def top_k_frequent(nums, k):
    freq = Counter(nums)
    heap = []
    for num, count in freq.items():
        heapq.heappush(heap, ([1], [2]))
    result = []
    for _ in range(k):
        freq, num = heapq.heappop(heap)
        result.append([3])
    return result

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

A-count

Bnum

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Using count instead of -count breaks max heap logic.

Appending freq instead of num adds counts, not elements.

Push (-count, num) to heap, pop and append num to result.

Practice

1. What is the main purpose of using a heap in the Top K Frequent Elements problem?

easy

A. To efficiently keep track of the top k elements by frequency

B. To sort all elements in ascending order

C. To store elements without any order

D. To remove duplicates from the list

Solution

1. Step 1: Understand the problem goal

   The problem asks to find the k elements that appear most frequently in a list.

2. Step 2: Role of heap in this context

   A heap helps efficiently keep track of the top k frequent elements by maintaining a small structure of size k, allowing quick access to the smallest frequency among them.

3. Final Answer:

   To efficiently keep track of the top k elements by frequency -> Option A

4. Quick Check:

   Heap = Efficient top k tracking [OK]

Hint: Heap keeps smallest frequency on top for quick replacement [OK]

Common Mistakes:

• Thinking heap sorts all elements fully
• Confusing heap with simple list or set
• Assuming heap removes duplicates

2. Which of the following Python code snippets correctly creates a min-heap from a list freq_list using the heapq module?

easy

A. import heapq
heapq.makeheap(freq_list)

B. import heapq
heapq.heapify(freq_list)

C. import heapq
heapq.heap(freq_list)

D. import heapq
heapq.create(freq_list)

Solution

1. Step 1: Recall heapq module functions

   The heapq module provides heapify() to convert a list into a heap in-place.

2. Step 2: Identify correct function usage

   Only heapq.heapify(freq_list) is valid syntax to create a min-heap from a list.

3. Final Answer:

   import heapq heapq.heapify(freq_list) -> Option B

4. Quick Check:

   heapify() creates heap [OK]

Hint: Use heapq.heapify() to build a heap from list [OK]

Common Mistakes:

• Using non-existent heapq functions
• Confusing heapify with sorting
• Missing import statement

3. What will be the output of the following code?

import heapq
nums = [1,1,1,2,2,3]
count = {}
for n in nums:
    count[n] = count.get(n, 0) + 1
heap = []
for num, freq in count.items():
    heapq.heappush(heap, (freq, num))
result = []
for _ in range(2):
    result.append(heapq.heappop(heap)[1])
print(result)

medium

A. [1, 2]

B. [2, 3]

C. [1, 3]

D. [3, 2]

Solution

1. Step 1: Count frequencies of elements

   nums = [1,1,1,2,2,3] gives count = {1:3, 2:2, 3:1}.

2. Step 2: Push (freq, num) into min-heap and pop twice

   Heap contains (3,1), (2,2), (1,3). Popping twice removes smallest freq first: (1,3), then (2,2). So result = [3, 2].

3. Final Answer:

   [3, 2] -> Option D

4. Quick Check:

   Lowest freq elements popped first = [3, 2] [OK]

Hint: Min-heap pops smallest frequency first [OK]

Common Mistakes:

• Assuming heap pops highest frequency first
• Mixing up element and frequency order
• Not counting frequencies correctly

4. Identify the error in the following code snippet that tries to find the top 2 frequent elements:

import heapq
nums = [4,4,4,5,5,6]
count = {}
for n in nums:
    count[n] = count.get(n, 0) + 1
heap = []
for num, freq in count.items():
    heapq.heappush(heap, (num, freq))
result = []
for _ in range(2):
    result.append(heapq.heappop(heap)[1])
print(result)

medium

A. The range in the last loop should be 3 instead of 2

B. The count dictionary is not updated correctly

C. The tuple pushed to heap should be (freq, num), not (num, freq)

D. heapq.heappush is used incorrectly; should use heapq.push

Solution

1. Step 1: Check tuple order pushed to heap

   The heap is a min-heap by first tuple element. Pushing (num, freq) orders by num, not frequency.

2. Step 2: Correct tuple order for frequency sorting

   To get top frequent elements, push (freq, num) so heap orders by frequency.

3. Final Answer:

   The tuple pushed to heap should be (freq, num), not (num, freq) -> Option C

4. Quick Check:

   Heap order depends on first tuple element [OK]

Hint: Push (frequency, element) tuples to heap [OK]

Common Mistakes:

• Swapping frequency and element in tuple
• Using wrong heapq function name
• Incorrect loop range for popping

5. Given the list nums = [2,3,2,4,3,3,5,5,5,5] and k = 2, which of the following code snippets correctly returns the top k frequent elements using a heap?

hard

A. import heapq
count = {}
for n in nums:
count[n] = count.get(n, 0) + 1
heap = []
for num, freq in count.items():
if len(heap) < k:
heapq.heappush(heap, (freq, num))
else:
if freq > heap[0][0]:
heapq.heapreplace(heap, (freq, num))
result = [num for freq, num in heap]
print(result)

B. import heapq
count = {}
for n in nums:
count[n] = count.get(n, 0) + 1
heap = []
for num, freq in count.items():
heapq.heappush(heap, (num, freq))
result = [heapq.heappop(heap)[1] for _ in range(k)]
print(result)

C. import heapq
count = {}
for n in nums:
count[n] = count.get(n, 0) + 1
heap = []
for num, freq in count.items():
heapq.heappush(heap, (-freq, num))
result = [heapq.heappop(heap)[0] for _ in range(k)]
print(result)

D. import heapq
count = {}
for n in nums:
count[n] = count.get(n, 0) + 1
heap = []
for num, freq in count.items():
if freq < k:
heapq.heappush(heap, (freq, num))
result = [num for freq, num in heap]
print(result)

Solution

1. Step 1: Understand the problem and approach

   We want top k frequent elements using a min-heap of size k to keep track of highest frequencies.

2. Step 2: Analyze each option

   import heapq count = {} for n in nums: count[n] = count.get(n, 0) + 1 heap = [] for num, freq in count.items(): if len(heap) < k: heapq.heappush(heap, (freq, num)) else: if freq > heap[0][0]: heapq.heapreplace(heap, (freq, num)) result = [num for freq, num in heap] print(result) correctly pushes (freq, num) and maintains heap size k, replacing smallest freq when a bigger freq is found. import heapq count = {} for n in nums: count[n] = count.get(n, 0) + 1 heap = [] for num, freq in count.items(): heapq.heappush(heap, (num, freq)) result = [heapq.heappop(heap)[1] for _ in range(k)] print(result) pushes (num, freq) which is wrong order. import heapq count = {} for n in nums: count[n] = count.get(n, 0) + 1 heap = [] for num, freq in count.items(): heapq.heappush(heap, (-freq, num)) result = [heapq.heappop(heap)[1] for _ in range(k)] print(result) uses negative freq but does not maintain heap size k properly. import heapq count = {} for n in nums: count[n] = count.get(n, 0) + 1 heap = [] for num, freq in count.items(): if freq < k: heapq.heappush(heap, (freq, num)) result = [num for freq, num in heap] print(result) pushes only if freq < k which is incorrect logic.

3. Final Answer:

   Option A code correctly returns top k frequent elements -> Option A

4. Quick Check:

   Min-heap size k with freq as key [OK]

Hint: Maintain min-heap size k with (freq, num) tuples [OK]

Common Mistakes:

• Pushing wrong tuple order to heap
• Not limiting heap size to k
• Using negative frequencies without proper logic