LeetCode in Kotlin
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
class Solution {
fun mostFrequentEven(nums: IntArray): Int {
val hm = HashMap<Int, Int>()
var max = 0
var small = Int.MAX_VALUE
if (nums.size == 1) {
return if (nums[0] % 2 == 0) nums[0]
else -1
}
for (i in nums.indices) {
if (nums[i] % 2 == 0) {
hm[nums[i]] = hm.getOrDefault(nums[i], 0) + 1
if (hm[nums[i]]!! > max) {
max = hm[nums[i]]!!
}
}
}
for ((key, value) in hm) {
if (value == max && key < small) {
small = key
}
}
return if (small == Int.MAX_VALUE) -1 else small
}
}