Most Frequent Even Element
<p>Given an integer array <code>nums</code>, return <em>the most frequent even element</em>.</p>

<p>If there is a tie, return the <strong>smallest</strong> one. If there is no such element, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,1,2,2,4,4,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong>
The even elements are 0, 2, and 4. Of these, 2 and 4 appear the most.
We return the smallest one, which is 2.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,4,4,9,2,4]
<strong>Output:</strong> 4
<strong>Explanation:</strong> 4 is the even element appears the most.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [29,47,21,41,13,37,25,7]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There is no even element.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>
Array; Hash Table; Counting
class Solution {
public:
    int mostFrequentEven(vector<int>& nums) {
        unordered_map<int, int> cnt;
        for (int x : nums) {
            if (x % 2 == 0) {
                ++cnt[x];
            }
        }
        int ans = -1, mx = 0;
        for (auto& [x, v] : cnt) {
            if (mx < v || (mx == v && ans > x)) {
                ans = x;
                mx = v;
            }
        }
        return ans;
    }
};