# Solving Leetcode Interviews in Seconds with AI: Most Frequent Even Element


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2404" 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, return the most frequent even element. If there is a tie, return the smallest one. If there is no such element, return -1.   Example 1:  Input: nums = [0,1,2,2,4,4,1] Output: 2 Explanation: The even elements are 0, 2, and 4. Of these, 2 and 4 appear the most. We return the smallest one, which is 2. Example 2:  Input: nums = [4,4,4,9,2,4] Output: 4 Explanation: 4 is the even element appears the most.  Example 3:  Input: nums = [29,47,21,41,13,37,25,7] Output: -1 Explanation: There is no even element.    Constraints:  1 <= nums.length <= 2000 0 <= nums[i] <= 105  

	# Explanation
	Here's the approach, complexity analysis, and Python code for finding the most frequent even element:

*   **Frequency Counting:** Use a dictionary (or hash map) to store the frequency of each even number in the input array.
*   **Finding the Maximum:** Iterate through the frequency map to identify the even number(s) with the highest frequency.
*   **Handling Ties:** If multiple even numbers have the same maximum frequency, return the smallest among them.

*   **Time Complexity:** O(n), where n is the length of the input array.
    **Space Complexity:** O(k), where k is the number of distinct even elements in the input array. In the worst case where all elements are even and distinct, k could be equal to n.

	
	# Code
	```python
	def mostFrequentEven(nums):
    """
    Finds the most frequent even element in an array.

    Args:
        nums: An integer array.

    Returns:
        The most frequent even element. If there is a tie, return the smallest one.
        If there is no such element, return -1.
    """

    even_counts = {}
    for num in nums:
        if num % 2 == 0:
            even_counts[num] = even_counts.get(num, 0) + 1

    if not even_counts:
        return -1

    max_frequency = 0
    most_frequent_even = -1

    for num, frequency in even_counts.items():
        if frequency > max_frequency:
            max_frequency = frequency
            most_frequent_even = num
        elif frequency == max_frequency:
            most_frequent_even = min(most_frequent_even, num)

    return most_frequent_even
	```