Most Frequent Even Element

Problem Statement

Problem 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.

Examples

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 of Given Problem

Intuition and Approach

The problem can be solved using a brute force approach or an optimized Technique.

• Brute Force
• Optimized approach

Approach 1:Brute Force (Naive)

Brute Force Approach: Iterate through the array and count the frequency of each even number using a dictionary or a similar data structure. Iterate through the frequency map to find the even number with the highest frequency. In case of a tie, select the smallest even number.

Codes in Different Languages

C++

Written by @AmruthaPariprolu

#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;

int mostFrequentEven(vector<int>& nums) {
    unordered_map<int, int> freq;
    for (int num : nums) {
        if (num % 2 == 0) {
            freq[num]++;
        }
    }

    int maxFreq = 0;
    int res = -1;
    for (auto& [num, count] : freq) {
        if (count > maxFreq || (count == maxFreq && num < res)) {
            maxFreq = count;
            res = num;
        }
    }

    return res;
}

int main() {
    vector<int> nums = {0, 1, 2, 2, 4, 4, 1};
    cout << mostFrequentEven(nums) << endl; // Output: 2
    return 0;
}

Java

Written by @AmruthaPariprolu

import java.util.HashMap;

public class Main {
    public static int mostFrequentEven(int[] nums) {
        HashMap<Integer, Integer> freq = new HashMap<>();
        for (int num : nums) {
            if (num % 2 == 0) {
                freq.put(num, freq.getOrDefault(num, 0) + 1);
            }
        }

        int maxFreq = 0;
        int res = -1;
        for (int num : freq.keySet()) {
            int count = freq.get(num);
            if (count > maxFreq || (count == maxFreq && num < res)) {
                maxFreq = count;
                res = num;
            }
        }

        return res;
    }

    public static void main(String[] args) {
        int[] nums = {0, 1, 2, 2, 4, 4, 1};
        System.out.println(mostFrequentEven(nums)); // Output: 2
    }
}

 

Python

Written by @AmruthaPariprolu

from collections import defaultdict

def most_frequent_even(nums):
    freq = defaultdict(int)
    for num in nums:
        if num % 2 == 0:
            freq[num] += 1

    max_freq = 0
    result = -1
    for num, count in freq.items():
        if count > max_freq or (count == max_freq and num < result):
            max_freq = count
            result = num

    return result

nums = [0, 1, 2, 2, 4, 4, 1]
print(most_frequent_even(nums))  # Output: 2

Complexity Analysis

• Time Complexity: $O(n^2)$
• due to the nested loop for counting frequency and finding the maximum.
• Space Complexity: $O(n)$
• for storing the frequency of numbers.

Video Explanation of Given Problem

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Authors:

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