<p>Given an array of integers <code>arr</code>, a <strong>lucky integer</strong> is an integer that has a frequency in the array equal to its value.</p>

<p>Return <em>the largest <strong>lucky integer</strong> in the array</em>. If there is no <strong>lucky integer</strong> return <code>-1</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [2,2,3,4]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The only lucky number in the array is 2 because frequency[2] == 2.
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,2,2,3,3,3]
<strong>Output:</strong> 3
<strong>Explanation:</strong> 1, 2 and 3 are all lucky numbers, return the largest of them.
</pre>

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

<pre>
<strong>Input:</strong> arr = [2,2,2,3,3]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There are no lucky numbers in the array.
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 500</code></li>
	<li><code>1 &lt;= arr[i] &lt;= 500</code></li>
</ul>

class Solution {
public:
    int findLucky(vector<int>& arr) {
        const int n = arr.size();
        sort(arr.begin(), arr.end());

        int i = n-1;
        while(i >= 0) {
            int cnt = 0;
            while(i-cnt >= 0 && arr[i-cnt] == arr[i]) ++cnt;

            if(cnt == arr[i]) return cnt;
            
            i -= cnt;
        }

        return -1;
    }
};
13100