Leetcode 2404 Most Frequent Even Element Solution in Java | Hindi Coding Community

 

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

Java Code :

class Solution {

public int mostFrequentEven(int[] A) {

HashMap<Integer,Integer> mp= new HashMap<>();

int val=1000000,freq=0;

for(var i:A){

if(i%2==0){

int curr= mp.getOrDefault(i,0)+1;

mp.put(i,curr);

if(curr>freq || curr==freq && i<val){

val=i;

freq=curr;

}

}

}

return freq==0? -1 : val;

}

}