README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2400-2499/2404.Most%20Frequent%20Even%20Element/README_EN.md	1259	Weekly Contest 310 Q1	Array Hash Table Counting

2404. Most Frequent Even Element

中文文档

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.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

Rust

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
    }
}

PHP

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;
    }
}