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).

2404 - Most Frequent Even Element

2404. Most Frequent Even Element

Problem

Solution (Java, C++, Python)

All Problems

All Solutions

Welcome to Subscribe On Youtube

2404. Most Frequent Even Element

Problem

Solution (Java, C++, Python)

All Problems

All Solutions