2404. Most Frequent Even Element LeetCode Solution

In this guide, you will get 2404. Most Frequent Even Element LeetCode Solution with the best time and space complexity. The solution to Most Frequent Even Element problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Most Frequent Even Element solution in C++
Most Frequent Even Element solution in Java
Most Frequent Even Element solution in Python
Additional Resources

2404. Most Frequent Even Element LeetCode Solution image

Problem Statement of Most Frequent Even Element

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

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

2404. Most Frequent Even Element LeetCode Solution in C++

class Solution {
 public:
  int mostFrequentEven(vector<int>& nums) {
    int ans = -1;
    unordered_map<int, int> count;

    for (const int num : nums) {
      if (num % 2 == 1)
        continue;
      const int newCount = ++count[num];
      const int maxCount = count[ans];
      if (newCount > maxCount || newCount == maxCount && num < ans)
        ans = num;
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

2404. Most Frequent Even Element LeetCode Solution in Java

class Solution {
  public int mostFrequentEven(int[] nums) {
    int ans = -1;
    Map<Integer, Integer> count = new HashMap<>();

    for (final int num : nums) {
      if (num % 2 == 1)
        continue;
      final int newCount = count.merge(num, 1, Integer::sum);
      final int maxCount = count.getOrDefault(ans, 0);
      if (newCount > maxCount || newCount == maxCount && num < ans)
        ans = num;
    }

    return ans;
  }
}
// code provided by PROGIEZ

2404. Most Frequent Even Element LeetCode Solution in Python

class Solution:
  def mostFrequentEven(self, nums: list[int]) -> int:
    ans = -1
    count = collections.Counter()

    for num in nums:
      if num % 2 == 1:
        continue
      count[num] += 1
      newCount = count[num]
      maxCount = count[ans]
      if newCount > maxCount or newCount == maxCount and num < ans:
        ans = num

    return ans
 # code by PROGIEZ