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Formatted question description: https://leetcode.ca/all/2404.html

# 2404. Most Frequent Even Element

• Difficulty: Easy.
• Related Topics: .
• Similar Questions: Majority Element, Majority Element II, Top K Frequent Elements, Sort Characters By Frequency.

## Problem

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

## Solution (Java, C++, Python)

• class Solution {
public int mostFrequentEven(int[] A) {
HashMap<Integer,Integer> mp= new HashMap<>();
int val=1000000,freq=0;
for(var i:A){
//if even element
if(i%2==0){
//increase frequency in map
int curr= mp.getOrDefault(i,0)+1;
mp.put(i,curr);
//Update smallest with greatest frequency
if(curr>freq || curr==freq && i<val){
val=i;
freq=curr;
}
}
}
return freq==0? -1 : val;
}
}

############

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

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

############

# 2404. Most Frequent Even Element
# https://leetcode.com/problems/most-frequent-even-element/

class Solution:
def mostFrequentEven(self, nums: List[int]) -> int:
counter = Counter()
have = False

for x in nums:
if x % 2 == 0:
counter[x] += 1
have = True

if not have: return -1

mmax = max(counter.values())

for x in sorted(counter.keys()):
if counter[x] == mmax:
return x

return -1


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

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

• function mostFrequentEven(nums: number[]): number {
const map = new Map();
for (const num of nums) {
if (num % 2 === 0) {
map.set(num, (map.get(num) ?? 0) + 1);
}
}
if (map.size === 0) {
return -1;
}

let res = 0;
let max = 0;
for (const [k, v] of map.entries()) {
if (v > max || (v == max && k < res)) {
max = v;
res = k;
}
}
return res;
}


• use std::collections::HashMap;
impl Solution {
pub fn most_frequent_even(nums: Vec<i32>) -> i32 {
let mut map = HashMap::new();
for &num in nums.iter() {
if num % 2 == 0 {
*map.entry(num).or_insert(0) += 1;
}
}
if map.len() == 0 {
return -1;
}

let mut res = 0;
let mut max = 0;
for (&k, &v) in map.iter() {
if v > max || (v == max && k < res) {
max = v;
res = k;
}
}
res
}
}


• 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;
}
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).