Top K Frequent Elements Using Heap in DSA C# - Time & Space Complexity

Time Complexity: Top K Frequent Elements Using Heap

O(n + m log k)

Understanding Time Complexity

We want to understand how the time needed to find the top K frequent elements changes as the input grows.

How does the program's speed change when the list of numbers gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

var frequency = new Dictionary<int, int>();
foreach (var num in nums) {
    frequency[num] = frequency.TryGetValue(num, out int cnt) ? cnt + 1 : 1;
}
var heap = new PriorityQueue<int, int>();
foreach (var kvp in frequency) {
    heap.Enqueue(kvp.Key, -kvp.Value);
    if (heap.Count > k) heap.Dequeue();
}
var result = new List<int>();
while (heap.Count > 0) result.Add(heap.Dequeue());
return result;
    

This code counts how often each number appears, then uses a heap to keep track of the top K frequent numbers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through the input array to count frequencies and looping through the frequency dictionary to manage the heap.
• How many times: The first loop runs once for each element in the input (n times). The second loop runs once for each unique number (m times).

How Execution Grows With Input

As the input size grows, counting frequencies grows linearly with n. Managing the heap grows with m and k.

Input Size (n)	Approx. Operations
10	About 10 to count + heap operations on unique elements
100	About 100 to count + heap operations on unique elements
1000	About 1000 to count + heap operations on unique elements

Pattern observation: Counting grows directly with input size, heap operations grow with unique elements and k, usually smaller than n.

Final Time Complexity

Time Complexity: O(n + m log k)

This means the time grows mostly with the input size n, but the heap operations depend on k, making it efficient when k is small.

Common Mistake

[X] Wrong: "The heap operations take O(n log n) because the heap size grows with n."

[OK] Correct: The heap size is kept at most k, so heap operations take O(log k), not O(log n), which is much faster when k is small.

Interview Connect

Understanding how to use heaps to efficiently find top K elements is a valuable skill that shows you can handle large data smartly and quickly.

Self-Check

"What if we used a simple sort on all unique elements instead of a heap? How would the time complexity change?"

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