# 2404. Most Frequent Even Element

## Description

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

## Solutions

Solution 1: Hash Table

We use a hash table $cnt$ to count the occurrence of all even elements, and then find the even element with the highest occurrence and the smallest value.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.

• class Solution {
public int mostFrequentEven(int[] nums) {
Map<Integer, Integer> cnt = new HashMap<>();
for (int x : nums) {
if (x % 2 == 0) {
cnt.merge(x, 1, Integer::sum);
}
}
int ans = -1, mx = 0;
for (var e : cnt.entrySet()) {
int x = e.getKey(), v = e.getValue();
if (mx < v || (mx == v && ans > x)) {
ans = x;
mx = v;
}
}
return ans;
}
}

• class Solution {
public:
int mostFrequentEven(vector<int>& nums) {
unordered_map<int, int> cnt;
for (int x : nums) {
if (x % 2 == 0) {
++cnt[x];
}
}
int ans = -1, mx = 0;
for (auto& [x, v] : cnt) {
if (mx < v || (mx == v && ans > x)) {
ans = x;
mx = v;
}
}
return ans;
}
};

• class Solution:
def mostFrequentEven(self, nums: List[int]) -> int:
cnt = Counter(x for x in nums if x % 2 == 0)
ans, mx = -1, 0
for x, v in cnt.items():
if v > mx or (v == mx and ans > x):
ans, mx = x, v
return ans


• func mostFrequentEven(nums []int) int {
cnt := map[int]int{}
for _, x := range nums {
if x%2 == 0 {
cnt[x]++
}
}
ans, mx := -1, 0
for x, v := range cnt {
if mx < v || (mx == v && x < ans) {
ans, mx = x, v
}
}
return ans
}

• function mostFrequentEven(nums: number[]): number {
const cnt: Map<number, number> = new Map();
for (const x of nums) {
if (x % 2 === 0) {
cnt.set(x, (cnt.get(x) ?? 0) + 1);
}
}
let ans = -1;
let mx = 0;
for (const [x, v] of cnt) {
if (mx < v || (mx === v && ans > x)) {
ans = x;
mx = v;
}
}
return ans;
}


• class Solution {
/**
* @param Integer[] $nums * @return Integer */ function mostFrequentEven($nums) {
$max =$rs = -1;
for ($i = 0;$i < count($nums);$i++) {
if ($nums[$i] % 2 == 0) {
$hashtable[$nums[$i]] += 1; if ($hashtable[$nums[$i]] > $max || ($hashtable[$nums[$i]] == $max &&$rs > $nums[$i])
) {
$max =$hashtable[$nums[$i]];
$rs =$nums[$i]; } } } return$rs;
}
}


• use std::collections::HashMap;
impl Solution {
pub fn most_frequent_even(nums: Vec<i32>) -> i32 {
let mut cnt = HashMap::new();
for &x in nums.iter() {
if x % 2 == 0 {
*cnt.entry(x).or_insert(0) += 1;
}
}
let mut ans = -1;
let mut mx = 0;
for (&x, &v) in cnt.iter() {
if mx < v || (mx == v && ans > x) {
ans = x;
mx = v;
}
}
ans
}
}