Leetcode 347: Top K Frequent Elements - DSA Interview Preparation

In this blog, we will explore how to solve the Leetcode problem '347. Top K Frequent Elements.' We'll begin by understanding the problem, move through a common brute force approach, provide a hint to guide you towards the efficient solution, and finally explain the efficient solution in detail.

Understanding the Problem Statement

The problem requires finding the k most frequent elements in a given list of integers. Given an integer array nums and an integer k, you need to return the k elements that appear most frequently. If there are multiple valid answers, any of them can be returned.

For example:

Input: nums = [1, 1, 1, 2, 2, 3], k = 2
Output: [1, 2]

The goal is to find the elements that occur most frequently in the given list and return exactly k such elements.

Brute Force Approach

A common brute force approach to solving this problem would be:

Count the Frequency: Iterate through the list and count the frequency of each element. You could use a dictionary or a list to store the counts of each unique element.
Sort by Frequency: After counting, you would need to sort the unique elements by their frequency in descending order.
Select Top K Elements: Finally, select the top k elements from the sorted list.

This brute force method is simple but inefficient for large lists. Sorting the entire list based on frequency takes O(n log n) time, where n is the number of unique elements. In addition, counting the frequency takes O(n) time, making the entire approach fairly slow for large datasets.

Hint to Solve the Problem Efficiently

To solve the problem more efficiently, consider using a heap data structure. The idea is to maintain the top k frequent elements without sorting the entire frequency list, which is where heaps come in handy.

The efficient solution consists of:

Using a dictionary to count the frequencies of each element.
Utilizing a heap to efficiently keep track of the k most frequent elements.

This approach helps reduce the time complexity significantly compared to the brute force approach.

Efficient Solution

Let's dive into the efficient solution, following the code provided:

from collections import Counter
import heapq

def topKFrequent(nums, k):
# Step 1: Count the frequency of each element
count = Counter(nums)  # O(n) time complexity

# Step 2: Use a heap to find the k most frequent elements
return heapq.nlargest(k, count.keys(), key=count.get)  # O(n log k) time complexity

Step-by-Step Explanation:

Count Frequencies: We use Python's Counter from the collections module to count the occurrences of each element in the list nums. This step takes O(n) time, where n is the number of elements in the list.
```
 count = Counter(nums)  # count will be a dictionary-like object with frequencies
```
Use a Heap: After counting the frequencies, we use the heapq.nlargest() function to find the k most frequent elements. This function returns the k largest elements from the iterable based on a key function, which in this case is count.get to access the frequency of each element. Using a heap here allows us to maintain the top k elements efficiently without sorting the entire frequency list.
```
 return heapq.nlargest(k, count.keys(), key=count.get)
```
The use of heapq.nlargest() has a time complexity of O(n log k), which is much more efficient than sorting the entire list.

Time and Space Complexity

Time Complexity: The time complexity of the solution is O(n log k).
- Counting the frequencies takes O(n) time.
- Using the heap to extract the k most frequent elements takes O(n log k) time.
- Therefore, the overall time complexity is dominated by O(n log k).
Space Complexity: The space complexity is O(n). We require space for storing the frequency counts in a dictionary (Counter). The heap can take up to k space, but since k <= n, the space required is effectively O(n).

Conclusion

The efficient solution to finding the top k frequent elements leverages Python's Counter to count the occurrences of elements and heapq.nlargest() to retrieve the most frequent elements efficiently. This approach is particularly beneficial when dealing with larger datasets, as it reduces the need to sort the entire frequency dictionary.

Top K Frequent Elements (Leetcode #347)

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