# Solving Leetcode Interviews in Seconds with AI: Top K Frequent Elements


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "347" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# 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.   Example 1: Input: nums = [1,1,1,2,2,3], k = 2 Output: [1,2] Example 2: Input: nums = [1], k = 1 Output: [1]    Constraints:  1 <= nums.length <= 105 -104 <= nums[i] <= 104 k is in the range [1, the number of unique elements in the array]. It is guaranteed that the answer is unique.    Follow up: Your algorithm's time complexity must be better than O(n log n), where n is the array's size. 

	# Explanation
	Here's a breakdown of the approach and the Python code for finding the k most frequent elements in an array:

*   **Frequency Counting:** Use a hash map (dictionary in Python) to count the frequency of each element in the input array.
*   **Bucket Sort (Radix Sort Idea):** Create a "bucket" array (list in Python) where the index represents the frequency and the value at that index is a list of elements with that frequency. This avoids full sorting and is more efficient.
*   **Retrieve Top k:** Iterate through the bucket array in reverse order (highest frequency to lowest) and add elements to the result until we have k elements.

*   **Runtime Complexity:** O(n), **Storage Complexity:** O(n)

	
	# Code
	```python
	def topKFrequent(nums, k):
    """
    Finds the k most frequent elements in an array.

    Args:
        nums: The input integer array.
        k: The number of most frequent elements to return.

    Returns:
        A list of the k most frequent elements.
    """

    # 1. Frequency Counting
    freq_map = {}
    for num in nums:
        freq_map[num] = freq_map.get(num, 0) + 1

    # 2. Bucket Sort
    bucket = [[] for _ in range(len(nums) + 1)]  # Initialize empty buckets
    for num, freq in freq_map.items():
        bucket[freq].append(num)

    # 3. Retrieve Top k
    result = []
    for i in range(len(bucket) - 1, 0, -1):
        for num in bucket[i]:
            result.append(num)
            if len(result) == k:
                return result

    return result  # Should not reach here as the problem guarantees a unique answer
	```