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 and popping the least frequent 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

In the context of Top K Frequent Elements, why do we use a frequency map before using a heap?

The frequency map counts how many times each element appears, which helps us know the frequency to compare when pushing elements into the heap.

intermediate

What happens when the heap size exceeds K during the process of finding top K frequent elements?

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

beginner

How do you extract the final top K frequent elements after building the heap?

Pop all elements from the heap and collect their values; these are the top K frequent elements.

Which data structure is best to efficiently track the top K frequent elements?

A Linked list

B Queue

C Stack

D Min-heap

A min-heap keeps the smallest frequency element on top, allowing efficient removal when size exceeds K.

What is the first step before using a heap to find top K frequent elements?

A Build a frequency map

B Sort the array

C Create a linked list

D Reverse the array

We first count the frequency of each element using a frequency map.

If the heap size is larger than K, what should you do?

A Remove the element with the lowest frequency

B Add more elements

C Remove the element with the highest frequency

D Do nothing

Remove the element with the lowest frequency to keep only the top K frequent elements.

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

A Only element values

B Only frequencies

C Pairs of (frequency, element)

D Sorted elements

The heap stores pairs of (frequency, element) to compare frequencies.

After building the heap, how do you get the final answer?

A Sort the heap

B Pop all elements from the heap

C Count elements again

D Ignore the heap

Pop all elements from the heap to get the top K frequent elements.

Explain step-by-step how to find the top K frequent elements using a heap.

Think about counting first, then using a heap to keep track of top K.

Why is a min-heap preferred over a max-heap for the Top K Frequent Elements problem?

Consider which element you want to remove when heap grows too big.

Practice

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

easy

A. To efficiently find and maintain the top K frequent elements

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 role of heap in frequency problems

   A heap helps keep track of the most frequent elements efficiently by allowing quick access to the smallest or largest frequency element.

2. Step 2: Identify why heap is used for top K elements

   Using a min-heap of size K allows us to maintain only the top K frequent elements, removing less frequent ones as we go.

3. Final Answer:

   To efficiently find and maintain the top K frequent elements -> Option A

4. Quick Check:

   Heap = Efficient top K tracking [OK]

Hint: Heap keeps top K elements efficiently by frequency [OK]

Common Mistakes:

• Thinking heap sorts all elements fully
• Using heap to remove duplicates
• Ignoring frequency counts in heap

2. Which of the following is the correct way to declare a min-heap in C++ using the standard library for storing pairs of (frequency, element)?

easy

A. priority_queue minHeap;

B. priority_queue> minHeap;

C. priority_queue, vector>, greater>> minHeap;

D. priority_queue, vector>, less>> minHeap;

Solution

1. Step 1: Understand C++ priority_queue syntax for min-heap

   By default, priority_queue is a max-heap. To make it min-heap, we use greater comparator.

2. Step 2: Check the correct template parameters

   We need a priority_queue of pairs, with vector as container, and greater> as comparator for min-heap.

3. Final Answer:

   priority_queue, vector>, greater>> minHeap; -> Option C

4. Quick Check:

   Min-heap = priority_queue with greater comparator [OK]

Hint: Use greater comparator for min-heap in priority_queue [OK]

Common Mistakes:

• Using default priority_queue which is max-heap
• Wrong comparator like less instead of greater
• Incorrect container type in priority_queue

3. What will be the output of the following code snippet that finds top 2 frequent elements?

#include 
#include 
#include 

int main() {
    int nums[] = {1,1,1,2,2,3};
    int freq[4] = {0};
    for(int i=0; i<6; i++) freq[nums[i]]++;
    int k = 2;
    int result[2];
    int idx = 0;
    for(int i=1; i<=3; i++) {
        if(idx < k) {
            result[idx++] = i;
        } else {
            int min_idx = 0;
            for(int j=1; j freq[result[min_idx]]) result[min_idx] = i;
        }
    }
    for(int i=0; i

medium

A. 3 2

B. 1 3

C. 2 3

D. 1 2

Solution

1. Step 1: Count frequencies of elements

   nums = {1,1,1,2,2,3} so freq[1]=3, freq[2]=2, freq[3]=1.

2. Step 2: Select top 2 frequent elements using simple logic

   Initially result = {1,2}. For i=3, freq[3]=1 is less than min freq in result (which is freq[2]=2), so no replacement.

3. Final Answer:

   1 2 -> Option D

4. Quick Check:

   Top 2 frequent = 1 and 2 [OK]

Hint: Count frequencies, keep top K by comparing min freq [OK]

Common Mistakes:

• Confusing element values with frequencies
• Replacing elements incorrectly in result array
• Ignoring initial filling of result array

4. Identify the bug in the following code snippet that tries to find top 3 frequent elements using a min-heap approach:

#include 
#include 
#include <queue>
#include <vector>
#include <unordered_map>
using namespace std;

int main() {
    vector<int> nums = {4,4,4,6,6,7,7,7,7};
    unordered_map freq;
    for(auto n : nums) freq[n]++;
    priority_queue, vector>, greater>> minHeap;
    int k = 3;
    for(auto it : freq) {
        minHeap.push({it.second, it.first});
        if(minHeap.size() > k) minHeap.pop();
    }
    while(!minHeap.empty()) {
        printf("%d ", minHeap.top().second);
        minHeap.pop();
    }
    return 0;
}

medium

A. The condition to pop from heap should be if size >= k, not > k

B. The output order is reversed; elements print from least to most frequent

C. The min-heap should store (element, frequency) pairs, not (frequency, element)

D. The frequency map is not initialized correctly

Solution

1. Step 1: Understand min-heap stores (frequency, element)

   Min-heap keeps smallest frequency on top, so popping prints from least to most frequent.

2. Step 2: Identify output order issue

   Printing while popping will print elements from lowest to highest frequency, not top frequent first.

3. Final Answer:

   The output order is reversed; elements print from least to most frequent -> Option B

4. Quick Check:

   Min-heap pop order = ascending frequency [OK]

Hint: Min-heap pops smallest first, so output order is reversed [OK]

Common Mistakes:

• Confusing pair order in heap
• Wrong pop condition for heap size
• Ignoring output order from heap pops

5. Given an array nums = {5,3,1,1,1,3,73,1} and k = 2, which approach correctly finds the top K frequent elements using a heap and handles ties properly?

hard

A. Use a min-heap of size k with (frequency, element) pairs, replacing when frequency is higher

B. Use a hash map only without any heap

C. Sort the array and pick the last k elements

D. Use a min-heap of (frequency, element) pairs and pop k times

Solution

1. Step 1: Count frequencies of elements

   Frequencies: 1->4, 3->2, 5->1, 73->1.

2. Step 2: Use min-heap of size k to keep top frequent elements

   Insert pairs (freq, element). If heap size exceeds k, pop smallest frequency. This keeps top k frequent elements efficiently.

3. Step 3: Handle ties by frequency naturally in heap

   Elements with same frequency are handled by heap order; ties do not break correctness.

4. Final Answer:

   Use a min-heap of size k with (frequency, element) pairs, replacing when frequency is higher -> Option A

5. Quick Check:

   Min-heap size k keeps top frequent elements [OK]

Hint: Min-heap size k keeps top frequent elements, replacing lower freq [OK]

Common Mistakes:

• Using max-heap and popping k times without size limit
• Sorting array instead of frequencies
• Ignoring heap for frequency tracking