Top K Frequent Elements Using Heap in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind using a heap to find the top K frequent elements?

Use a heap to keep track of the K elements with the highest frequency by pushing elements with their counts and popping the smallest frequency when size exceeds K.

beginner

What type of heap is commonly used to solve the Top K Frequent Elements problem efficiently?

A min-heap is used to keep the smallest frequency element at the top, so when the heap size exceeds K, the least frequent element can be removed.

beginner

Why do we use a hash map (unordered_map) along with a heap in this problem?

The hash map counts the frequency of each element quickly, and the heap helps select the top K elements based on these frequencies.

intermediate

In C++, which STL container is typically used to implement the min-heap for this problem?

std::priority_queue with a custom comparator or by storing pairs with frequency as the first element and using std::greater to make it a min-heap.

intermediate

What is the time complexity of finding the top K frequent elements using a heap?

O(N log K), where N is the number of unique elements, because each element is pushed and possibly popped from the heap of size K.

Which data structure is used to count the frequency of elements before using a heap?

A Queue

B Stack

C Hash map (unordered_map)

D Linked list

A hash map is used to count frequencies efficiently by mapping elements to their counts.

What does the heap store in the Top K Frequent Elements problem?

A Elements only

B Indices of elements

C Frequencies only

D Pairs of (frequency, element)

The heap stores pairs of (frequency, element) to compare frequencies and keep track of elements.

Why do we use a min-heap instead of a max-heap for this problem?

A To keep the smallest frequency at the top for easy removal

B Because max-heap is slower

C To sort elements alphabetically

D To store elements in ascending order

A min-heap keeps the smallest frequency on top so when the heap size exceeds K, the least frequent element can be removed.

What happens when the heap size exceeds K during insertion?

A The largest frequency element is removed

B The smallest frequency element is removed

C The heap is cleared

D No element is removed

The smallest frequency element is removed to keep only the top K frequent elements.

What is the overall time complexity of this approach?

A O(N log K)

B O(N log N)

C O(N)

D O(K log N)

Counting frequencies is O(N), and maintaining a heap of size K for N elements is O(N log K), so overall O(N log K).

Explain how to use a heap and a hash map together to find the top K frequent elements in a list.

Think about counting first, then selecting top K with a heap.

Describe the role of the min-heap in the Top K Frequent Elements algorithm and why it is preferred over a max-heap.

Consider what happens when you add more than K elements.

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 requires finding the k elements that appear most frequently.

2. Step 2: Role of heap in this context

   A heap helps efficiently keep track of the top k frequent elements by maintaining a structure where the smallest frequency is easily accessible and can be replaced if a more frequent element is found.

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 array storage
• Assuming heap removes duplicates

2. Which of the following C++ STL containers is best suited to implement a min-heap for the Top K Frequent Elements problem?

easy

A. std::vector with std::make_heap and std::pop_heap

B. std::list

C. std::map

D. std::queue

Solution

1. Step 1: Identify heap implementation in C++ STL

   C++ STL uses a vector with heap algorithms like std::make_heap, std::pop_heap, and std::push_heap to implement heaps.

2. Step 2: Check other container suitability

   std::list and std::queue do not support heap operations directly. std::map is a balanced tree, not a heap.

3. Final Answer:

   std::vector with std::make_heap and std::pop_heap -> Option A

4. Quick Check:

   Vector + heap algorithms = min-heap [OK]

Hint: Use vector + heap functions for heap in C++ [OK]

Common Mistakes:

• Choosing std::map thinking it is a heap
• Using std::queue which is FIFO, not heap
• Using std::list which lacks random access

3. What is the output of the following C++ code snippet for nums = {1,1,1,2,2,3} and k = 2?

std::vector<int> nums = {1,1,1,2,2,3};
int k = 2;
// Frequency map
std::unordered_map freq;
for (int n : nums) freq[n]++;

// Min heap to store pairs (frequency, element)
std::priority_queue, std::vector>, std::greater<>> minHeap;

for (auto& p : freq) {
    minHeap.push({p.second, p.first});
    if (minHeap.size() > k) minHeap.pop();
}

std::vector<int> result;
while (!minHeap.empty()) {
    result.push_back(minHeap.top().second);
    minHeap.pop();
}
std::reverse(result.begin(), result.end());
for (int x : result) std::cout << x << " ";

medium

A. 3 2

B. 2 3

C. 1 3

D. 1 2

Solution

1. Step 1: Calculate frequencies

   Frequencies: 1 -> 3 times, 2 -> 2 times, 3 -> 1 time.

2. Step 2: Use min heap to keep top 2 frequent elements

   Insert (3,1), (2,2), (1,3). When size exceeds 2, pop smallest frequency. After all insertions, heap contains (2,2) and (3,1).

3. Step 3: Extract and reverse result

   Extracting gives elements 2 and 1, reversed to 1 2.

4. Final Answer:

   1 2 -> Option D

5. Quick Check:

   Top 2 frequent = 1 and 2 [OK]

Hint: Heap keeps smallest freq; pop when size > k [OK]

Common Mistakes:

• Not reversing the result vector
• Confusing element and frequency in pairs
• Popping wrong element from heap

4. Identify the bug in this code snippet for finding top k frequent elements:

std::unordered_map freq;
for (int n : nums) freq[n]++;

std::priority_queue> maxHeap;
for (auto& p : freq) {
    maxHeap.push({p.second, p.first});
    if (maxHeap.size() > k) maxHeap.pop();
}

std::vector<int> result;
while (!maxHeap.empty()) {
    result.push_back(maxHeap.top().second);
    maxHeap.pop();
}

medium

A. Frequency map is not created correctly

B. Using maxHeap and popping when size > k removes most frequent elements

C. Result vector is not reversed before output

D. Priority queue should store only elements, not pairs

Solution

1. Step 1: Understand heap type and popping logic

   maxHeap stores pairs with highest frequency on top. Popping when size > k removes the top (most frequent) element, which is wrong.

2. Step 2: Correct approach

   Use a minHeap to keep smallest frequency on top and pop it when size > k, so top k frequent remain.

3. Final Answer:

   Using maxHeap and popping when size > k removes most frequent elements -> Option B

4. Quick Check:

   MaxHeap pop removes top freq, wrong for top k [OK]

Hint: Pop from minHeap, not maxHeap, to keep top k [OK]

Common Mistakes:

• Using maxHeap and popping top element
• Not reversing result vector if needed
• Ignoring frequency-element pair structure

5. Given a large input array with many elements having the same frequency, how can you modify the heap approach to ensure the Top K Frequent Elements always returns the smallest elements when frequencies tie? Choose the best approach.

hard

A. Use a hash map only without any heap

B. Use a max-heap ordered by frequency only

C. Use a min-heap ordered by frequency, then by element value descending

D. Sort the entire array before building the heap

Solution

1. Step 1: Understand tie-breaking requirement

   When frequencies tie, we want the smaller element to come first in the top k.

2. Step 2: Modify heap ordering

   Use a min-heap that orders pairs by frequency first, then by element value ascending to break ties correctly.

3. Step 3: Why other options fail

   Max-heap ignores tie-break order, sorting entire array is inefficient, and hash map alone cannot find top k efficiently.

4. Final Answer:

   Use a min-heap ordered by frequency, then by element value ascending -> Option C

5. Quick Check:

   Min-heap with tie-break by element value [OK]

Hint: Tie-break by element value in heap comparator [OK]

Common Mistakes:

• Ignoring tie-break condition in heap
• Sorting entire array wastes time
• Using hash map alone without heap