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
Related Topics:
Array, Hash Table, Counting
Similar Questions:
- Majority Element (Easy)
- Majority Element II (Medium)
- Top K Frequent Elements (Medium)
- Sort Characters By Frequency (Medium)
// OJ: https://leetcode.com/problems/most-frequent-even-element
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
int mostFrequentEven(vector<int>& A) {
unordered_map<int, int> cnt;
for (int n : A) {
if (n % 2 == 0) cnt[n]++;
}
int maxFreq = 0, ans = -1;
for (auto &[n, c] : cnt) {
if (c > maxFreq) {
ans = n;
maxFreq = c;
} else if (c == maxFreq && n < ans) ans = n;
}
return ans;
}
};