README.md

2404. 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

Solutions (Rust)

1. Solution

use std::collections::HashMap;

impl Solution {
    pub fn most_frequent_even(nums: Vec<i32>) -> i32 {
        let mut count = HashMap::new();
        let mut max_count = 0;
        let mut ret = -1;

        for num in nums.iter().filter(|&&x| x % 2 == 0) {
            count.entry(num).and_modify(|x| *x += 1).or_insert(1);
        }

        for (&k, v) in count {
            if v > max_count || (v == max_count && k < ret) {
                max_count = v;
                ret = k;
            }
        }

        ret
    }
}