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, 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

    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