What is the optimal time complexity of solving Top K Frequent Elements?

The optimized solution achieves a time complexity of O(n) by using sorting, mathematical shortcuts, or hash table lookups.

What is the auxiliary space complexity for Top K Frequent Elements?

The space complexity is O(n) since we only use a minimal/constant amount of extra memory for tracking variables (or auxiliary space if dynamic structures are allocated).

What are the typical edge cases we must handle in Top K Frequent Elements?

Key edge cases include empty collections, negative or large bounds causing integer overflow, arrays with only one element, or inputs where target criteria are not met.

How can we optimize the brute force approach for Top K Frequent Elements?

Brute force methods often inspect all possible combinations. We optimize this by sorting the input list to enable binary search, or tracking processed items in a set to avoid redundant nested loop lookups.

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Top K Frequent Elements

Learn how to solve the 'Top K Frequent Elements' problem. This detailed resource details brute force and optimized approaches.

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Problem Statement

Easy

Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.

Write a function topKFrequent(nums: List[int], k: int) -> List[int].

Constraints

•1 <= len(nums) <= 10^5
•-10^4 <= nums[i] <= 10^4
•k is in the range [1, number of unique elements in nums]
•The answer is guaranteed to be unique

Examples

Example 1

Input

nums = [1, 1, 1, 2, 2, 3], k = 2

Output

[1, 2]

Explanation

1 appears 3 times and 2 appears 2 times. These are the 2 most frequent elements.

Example 2

Input

nums = [1], k = 1

Output

[1]

Explanation

There is only one element, so it is the most frequent.

Need a Hint?

Analyze the input constraints. Try sorting first (O(n log n)) or using a hash map/set to track seen elements in O(n) time.

Edge Cases to Watch

Empty list or null input variables
Single item lists/arrays
Extremely large input bounds causing integer or stack overflow

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