Top K Frequent Elements
Detailed guide and Python implementation for the 'Top K Frequent Elements' problem.
1. Concept Overview
The 'Top K Frequent Elements' problem is a key challenge in the Arrays & Hashing section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Top K Frequent Elements.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Top K Frequent Elements carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
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].
- •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
nums = [1, 1, 1, 2, 2, 3], k = 2
[1, 2]
1 appears 3 times and 2 appears 2 times. These are the 2 most frequent elements.
nums = [1], k = 1
[1]
There is only one element, so it is the most frequent.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
Open the problem in PyRun's browser-based Python editor. Your code runs fully offline — no server required.
Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
import heapq
def top_k_frequent_opt(nums, k):
count = {}
for n in nums:
count[n] = 1 + count.get(n, 0)
return heapq.nlargest(k, count.keys(), key=count.get)Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def top_k_frequent_brute(nums, k):
count = {}
for n in nums:
count[n] = 1 + count.get(n, 0)
sorted_counts = sorted(count.items(), key=lambda x: x[1], reverse=True)
return [item[0] for item in sorted_counts[:k]]Algorithm Pattern Checklist
When dealing with Arrays & Hashing data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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