Top K Frequent Elements Using Heap in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a dictionary to count frequencies.

DSA C#

var frequencyMap = new Dictionary<int, [1]>();

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

Abool

Bstring

CList<int>

Dint

Attempts:

3 left

💡 Hint

Common Mistakes

Using string or bool as the value type instead of int.

The dictionary maps integers to their frequency counts, which are integers.

2fill in blank

medium

Complete the code to add or update frequency count in the dictionary.

DSA C#

if (frequencyMap.ContainsKey(num)) {
    frequencyMap[num] [1] 1;
} else {
    frequencyMap[num] = 1;
}

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

A+=

B-=

C*=

D/=

Attempts:

3 left

💡 Hint

Common Mistakes

Using -= or *= which would decrease or multiply the count incorrectly.

We add 1 to the existing frequency count when the number is already in the dictionary.

3fill in blank

hard

Fix the error in the code to create a min-heap using PriorityQueue.

DSA C#

var minHeap = new PriorityQueue<int, int>();
foreach (var entry in frequencyMap) {
    minHeap.Enqueue(entry.Key, [1]);
    if (minHeap.Count > k) {
        minHeap.Dequeue();
    }
}

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

Aentry.Key

Bentry.Value

Ck

DminHeap.Count

Attempts:

3 left

💡 Hint

Common Mistakes

Using entry.Key as priority which is the element, not frequency.

The priority for the heap should be the frequency (entry.Value) to keep the least frequent element on top.

4fill in blank

hard

Fill both blanks to extract elements from the heap into a list in correct order.

DSA C#

var result = new List<int>();
while (minHeap.Count > 0) {
    result.[1](minHeap.[2]());
}
result.Reverse();

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

AAdd

BDequeue

CPeek

DRemove

Attempts:

3 left

💡 Hint

Common Mistakes

Using Peek instead of Dequeue which does not remove the element.

Using Remove which is not a method of PriorityQueue.

We remove elements from the heap using Dequeue and add them to the result list using Add.

5fill in blank

hard

Fill all three blanks to complete the method returning top k frequent elements.

DSA C#

public IList<int> TopKFrequent(int[] nums, int k) {
    var frequencyMap = new Dictionary<int, int>();
    foreach (var num in nums) {
        if (frequencyMap.ContainsKey(num)) {
            frequencyMap[num] [1] 1;
        } else {
            frequencyMap[num] = 1;
        }
    }
    var minHeap = new PriorityQueue<int, int>();
    foreach (var entry in frequencyMap) {
        minHeap.Enqueue(entry.Key, entry.Value);
        if (minHeap.Count > k) {
            minHeap.[2]();
        }
    }
    var result = new List<int>();
    while (minHeap.Count > 0) {
        result.[3](minHeap.Dequeue());
    }
    result.Reverse();
    return result;
}

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

A+=

BDequeue

CAdd

DRemove

Attempts:

3 left

💡 Hint

Common Mistakes

Using Remove instead of Dequeue for heap removal.

Using Remove instead of Add for list insertion.

We increment frequency with +=, remove smallest frequency with Dequeue, and add elements to result with Add.

Practice

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

easy

A. To store elements without any order

B. To sort all elements in ascending order

C. To efficiently keep track of the top K elements by frequency

D. To remove duplicates from the list

Solution

1. Step 1: Understand the role of a heap

   A heap is a special tree-based data structure that helps efficiently find and maintain the largest or smallest elements.

2. Step 2: Apply heap to the problem context

   In the Top K Frequent Elements problem, a heap helps keep track of the K elements with the highest frequency without sorting the entire list.

3. Final Answer:

   To efficiently keep track of the top K elements by frequency -> Option C

4. Quick Check:

   Heap = Efficient top K tracking [OK]

Hint: Heap keeps top K elements efficiently without full sorting [OK]

Common Mistakes:

• Thinking heap sorts all elements fully
• Confusing heap with simple list or array
• Ignoring frequency importance

2. Which of the following C# code snippets correctly declares a min-heap using SortedSet for integers?

easy

A. var minHeap = new Stack<int>();

B. var minHeap = new List<int>();

C. var minHeap = new Dictionary();

D. var minHeap = new SortedSet<int>();

Solution

1. Step 1: Identify data structure for min-heap

   A min-heap can be implemented using a SortedSet in C# because it keeps elements sorted automatically.

2. Step 2: Check code correctness

   var minHeap = new SortedSet<int>(); declares a SortedSet<int>, which acts like a min-heap by keeping smallest element accessible.

3. Final Answer:

   var minHeap = new SortedSet<int>(); -> Option D

4. Quick Check:

   SortedSet = min-heap structure [OK]

Hint: Use SortedSet for min-heap in C# [OK]

Common Mistakes:

• Using List instead of a heap structure
• Confusing Dictionary with heap
• Using Stack which is LIFO, not heap

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

int[] nums = {1,1,1,2,2,3};
int k = 2;
var freq = new Dictionary();
foreach(var num in nums) freq[num] = freq.GetValueOrDefault(num, 0) + 1;
var heap = new SortedSet<(int freq, int num)>();
foreach(var pair in freq) {
    heap.Add((pair.Value, pair.Key));
    if(heap.Count > k) heap.Remove(heap.Min);
}
var result = heap.Select(x => x.num).ToList();
Console.WriteLine(string.Join(", ", result));

medium

A. 2, 1

B. 1, 2

C. 1, 3

D. 3, 2

Solution

1. Step 1: Count frequencies

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

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

   Heap keeps pairs (freq, num). It adds all pairs: (3,1), (2,2), (1,3). When size > 2, removes smallest freq (1,3). Leaving (2,2) and (3,1). SortedSet iterates in ascending order: (2,2), then (3,1).

3. Step 3: Output order

   Result list is created by selecting num from heap in ascending order of freq, so output is "2, 1".

4. Final Answer:

   2, 1 -> Option A

5. Quick Check:

   Top 2 frequent = 2, 1 [OK]

Hint: Keep heap size k, remove smallest freq when exceeded [OK]

Common Mistakes:

• Not removing smallest freq when heap size exceeds k
• Mixing element and frequency order
• Printing frequencies instead of elements

4. Identify the bug in this C# code snippet for top 1 frequent element:

int[] nums = {4,4,4,6,6,7};
int k = 1;
var freq = new Dictionary();
foreach(var num in nums) freq[num] = freq.GetValueOrDefault(num, 0) + 1;
var heap = new SortedSet<(int freq, int num)>();
foreach(var pair in freq) {
    heap.Add((pair.Value, pair.Key));
    if(heap.Count > k) heap.Remove(heap.Max);
}
var result = heap.Select(x => x.num).ToList();
Console.WriteLine(string.Join(", ", result));

medium

A. Removing heap.Max instead of heap.Min causes wrong elements removal

B. Dictionary initialization is incorrect

C. Using GetValueOrDefault causes runtime error

D. Heap should store only numbers, not frequency

Solution

1. Step 1: Understand heap removal logic

   Heap is a min-heap by frequency, so smallest freq is at Min, largest at Max.

2. Step 2: Identify incorrect removal

   Code removes heap.Max when size exceeds k, which removes the largest frequency element instead of smallest, causing wrong result.

3. Final Answer:

   Removing heap.Max instead of heap.Min causes wrong elements removal -> Option A

4. Quick Check:

   Remove heap.Min to keep top frequent [OK]

Hint: Remove heap.Min to keep top frequent elements [OK]

Common Mistakes:

• Removing heap.Max instead of heap.Min
• Confusing frequency and element order
• Not updating frequency dictionary correctly

5. Given the array nums = {5,3,1,1,1,3,73,1} and k = 2, which approach correctly returns the top 2 frequent elements using a heap in C#? A) Use a max-heap to store all elements and pop k times. B) Use a min-heap of size k to keep top frequent elements by frequency. C) Sort the array and pick last k elements. D) Use a queue to store elements and count frequencies later.

hard

A. Use a max-heap to store all elements and pop k times.

B. Use a min-heap of size k to keep top frequent elements by frequency.

C. Sort the array and pick last k elements.

D. Use a queue to store elements and count frequencies later.

Solution

1. Step 1: Understand frequency counting

   Count frequencies: 1 -> 4 times, 3 -> 2 times, 5 -> 1 time, 73 -> 1 time.

2. Step 2: Use min-heap of size k

   Use a min-heap to keep only top k frequent elements by frequency. When heap size exceeds k, remove smallest frequency element.

3. Step 3: Why other options fail

   A) Max-heap of all elements is inefficient. C) Sorting array doesn't group frequencies. D) Queue doesn't help frequency counting efficiently.

4. Final Answer:

   Use a min-heap of size k to keep top frequent elements by frequency. -> Option B

5. Quick Check:

   Min-heap size k = correct approach [OK]

Hint: Keep min-heap size k to track top frequent elements [OK]

Common Mistakes:

• Using max-heap for all elements wastes memory
• Sorting array ignores frequency counts
• Using queue without frequency map