Top K Frequent Elements Using Heap in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Top K Frequent Elements with Min-Heap

What is the output of the following code that finds the top 2 frequent elements using a min-heap?

DSA C

int nums[] = {1,1,1,2,2,3};
int k = 2;
// Frequency map: {1:3, 2:2, 3:1}
// Using a min-heap to keep top k frequent elements
// Final output is the elements in the heap
// Output order does not matter

// Result: [1, 2]

A [1, 2]

B [3, 2]

C [1, 3]

D [2, 3]

💡 Hint

Think about which elements appear most often and how a min-heap keeps the smallest frequency at the top.

The frequencies are 1:3, 2:2, 3:1. The top 2 frequent elements are 1 and 2. The min-heap keeps the smallest frequency on top, so 3 is removed first.

🧠 Conceptual

intermediate

Understanding Heap Usage in Top K Frequent Elements

Why do we use a min-heap instead of a max-heap when finding the top K frequent elements?

A Because a min-heap sorts elements in descending order automatically.

B Because a min-heap allows us to efficiently remove the least frequent element when the heap size exceeds k.

C Because a max-heap cannot store frequencies.

D Because a max-heap is slower than a min-heap.

💡 Hint

Think about how to keep only the top k elements in the heap.

Using a min-heap of size k lets us keep the k most frequent elements by removing the smallest frequency when adding a new element.

🔧 Debug

advanced

Identify the Error in Heap Implementation

What error will this code cause when trying to find the top 3 frequent elements? Code snippet: int nums[] = {4,4,4,6,6,7}; int k = 3; // Frequency map: {4:3, 6:2, 7:1} // Min-heap stores pairs (frequency, element) // Heap push: push(freq, element) // Heap pop: pop smallest frequency // Missing heapify after push // What error occurs?

A No error; code runs correctly.

B Syntax error due to missing semicolon.

C Runtime error due to null pointer.

D The heap will not maintain the correct order, leading to wrong output.

💡 Hint

Think about what happens if the heap property is not maintained after insertion.

Without heapify after push, the heap structure breaks, so the smallest frequency may not be at the top, causing wrong results.

❓ Predict Output

advanced

Output of Top K Frequent Elements with Equal Frequencies

What is the output of the following code for top 2 frequent elements? int nums[] = {5,5,6,6,7}; int k = 2; // Frequencies: {5:2, 6:2, 7:1} // Min-heap keeps top k elements by frequency // If frequencies tie, smaller element is preferred // What is the output?

A [5, 6]

B [6, 5]

C [5, 7]

D [6, 7]

💡 Hint

When frequencies tie, check which element is smaller.

Both 5 and 6 have frequency 2, but 5 is smaller, so the heap keeps 5 and 6 as top 2.

🚀 Application

expert

Heap Operations Sequence for Top K Frequent Elements

Given the array nums = {2,3,2,4,3,3,5} and k = 2, which sequence of heap operations correctly maintains the min-heap of size k for top frequent elements? Ordering targets: 1: push(1,5) 2: push(2,2) 3: push(3,3) 4: pop() Options: A: 2,3,1,4 B: 1,2,3,4 C: 2,1,3,4 D: 1,3,2,4

A 2,1,3,4

B 1,2,3,4

C 2,3,1,4

D 1,3,2,4

💡 Hint

Push elements in order of frequency, then pop the smallest frequency to keep heap size k.

Push frequency 2 for element 2, then frequency 3 for element 3, then frequency 1 for element 5. After pushing all, pop the smallest frequency (1,5) to keep size 2.

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