Solving Top K Frequent Elements in C# with PriorityQueue
1. The Problem: Top K Frequent Elements
The “Top K Frequent Elements” problem is a popular challenge that tests your knowledge of frequency counting and efficient sorting.
Given an integer array
numsand an integerk, return thekmost frequent elements. You may return the answer in any order.
2. The Intuition: Trading Speed for Space
The most straightforward way to solve this would be to count the frequencies and then sort the entire list of unique numbers. However, sorting takes O(N log N) time.
By using a Min-Heap (PriorityQueue), we can maintain only the “Top K” elements as we iterate, reducing our time complexity to O(N log K).
- Count Everything: Use a
Dictionaryto store how many times each number appears. - The “K-Sized Tray” (Min-Heap): We iterate through our counts. We add elements to a Min-Heap where the frequency is the priority.
- Automatic Filter: If the heap’s size exceeds
k, weDequeue()the element with the lowest frequency. - The Survivors: After checking all elements, the
kelements remaining in the heap are our most frequent ones.
3. Implementation (Version 1: PriorityQueue)
Here is a clean implementation using Dictionary<int, int> for counting and PriorityQueue<int, int> for maintaining the top elements.
public class Solution {
public int[] TopKFrequent(int[] nums, int k) {
// 1. Initialize our frequency map and the priority queue (Min-Heap)
var minHeap = new PriorityQueue<int, int>();
var freq = new Dictionary<int, int>();
// 2. Count the frequency of each number
foreach (var num in nums) {
freq[num] = freq.GetValueOrDefault(num, 0) + 1;
}
// 3. Iterate through the dictionary and maintain a heap of size k
foreach (var (num, count) in freq) {
// Enqueue the number with its frequency as the priority
minHeap.Enqueue(num, count);
// If we have more than k elements, remove the one with the lowest frequency
if (minHeap.Count > k) {
minHeap.Dequeue();
}
}
// 4. Extract the elements from the heap
// UnorderedItems allows us to access elements without clearing the queue
return minHeap.UnorderedItems.Select(x => x.Element).ToArray();
}
}
4. Implementation (Neetcode Version)
This version follows the Neetcode approach, which also uses a Min-Heap (PriorityQueue) but handles frequency counting and result extraction more manually.
public class Solution {
public int[] TopKFrequent(int[] nums, int k) {
// 1. Manually count frequencies using a Dictionary
var count = new Dictionary<int, int>();
foreach (var num in nums) {
if (count.ContainsKey(num)) {
count[num]++;
} else {
count[num] = 1;
}
}
// 2. Maintain a Min-Heap of size K
var heap = new PriorityQueue<int, int>();
foreach (var entry in count) {
heap.Enqueue(entry.Key, entry.Value);
if (heap.Count > k) {
heap.Dequeue();
}
}
// 3. Extract the K survivors from the heap
var res = new int[k];
for (int i = 0; i < k; i++) {
res[i] = heap.Dequeue();
}
return res;
}
}
5. Step-by-Step Breakdown
Step 1: Frequency Count
We use a Dictionary<int, int> where the Key is the number and the Value is its frequency. GetValueOrDefault makes the increment logic concise.
Step 2: The Min-Heap Strategy
We use the PriorityQueue<TElement, TPriority> introduced in .NET 6. By using the frequency as the priority, the “smallest” (least frequent) item is always at the top of the heap.
Step 3: Maintaining Size K
Every time the heap grows beyond size k, we call Dequeue(). This removes the element with the lowest frequency currently in the heap. This ensures that only the k largest (most frequent) elements remain.
Step 4: Efficient Extraction
Instead of manually dequeuing elements into a list, we use .UnorderedItems. This property provides a non-destructive way to access the elements in the queue, which we then project and convert to an array.
6. Complexity Analysis
| Metric | Complexity | Why? |
|---|---|---|
| Time Complexity | O(N log K) | We iterate over N elements to build the dictionary (O(N)). Then we iterate over unique elements (max N) and perform heap operations (log K). |
| Space Complexity | O(N) | In the worst case (all unique elements), the dictionary stores N items. The heap stores K items. |
7. Why use a PriorityQueue?
If we used Array.Sort() on the dictionary entries, it would be O(N log N).
- For $N = 1,000,000$ and $k = 10$:
- Sorting: ~20,000,000 operations.
- PriorityQueue: ~3,300,000 operations.
The PriorityQueue approach is significantly faster when k is small compared to N.
8. Summary
The Top K Frequent Elements problem demonstrates how to combine two fundamental data structures: Dictionaries for fast counting and Heaps for maintaining a rolling “Top List” of items.
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