LeetCode-in-Java
2404. Most Frequent Even Element
Easy
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 <= 20000 <= nums[i] <= 105
Solution
import java.util.HashMap;
import java.util.Map;
public class Solution {
public int mostFrequentEven(int[] nums) {
Map<Integer, Integer> frequencyMap = new HashMap<>();
int mostFrequent = Integer.MAX_VALUE;
int maxFrequency = 0;
for (int num : nums) {
if ((num & 1) == 0) {
maxFrequency =
Math.max(
maxFrequency,
frequencyMap.compute(num, (n, freq) -> freq == null ? 1 : ++freq));
}
}
for (Map.Entry<Integer, Integer> entry : frequencyMap.entrySet()) {
if (entry.getValue() == maxFrequency) {
mostFrequent = Math.min(mostFrequent, entry.getKey());
}
}
return mostFrequent == Integer.MAX_VALUE ? -1 : mostFrequent;
}
}