<p>You are given an integer array <code>arr</code>. You can choose a set of integers and remove all the occurrences of these integers in the array.</p>

<p>Return <em>the minimum size of the set so that <strong>at least</strong> half of the integers of the array are removed</em>.</p>

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

<pre>
<strong>Input:</strong> arr = [3,3,3,3,5,5,5,2,2,7]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has a size greater than half of the size of the old array.
</pre>

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

<pre>
<strong>Input:</strong> arr = [7,7,7,7,7,7]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The only possible set you can choose is {7}. This will make the new array empty.
</pre>

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

<ul>
	<li><code>2 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
	<li><code>arr.length</code> is even.</li>
	<li><code>1 &lt;= arr[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    int minSetSize(vector<int>& arr) {
        int ans = 0;
        // count frequency of ints
        int m = 0;
        unordered_map<int, int> F;
        for (int &n: arr) {
            m = max(m, ++F[n]);
        }
        // count freqnecy of frequencies
        int n = 0;
        int *FF = new int[++m];
        memset(FF, 0, sizeof(int) * m);
        for (auto &p: F) {
            ++FF[p.second];
            n = max(m, p.second);
        }

        // grab most frequent ints and remove those first
        int o = (arr.size() + 1) / 2;
        for (int i = n-1; i > 0 && o > 0; --i) {
            while (i > 0 && !FF[i]) --i;
            if (o >= FF[i] * i) {
                o -= FF[i] * i;
                ans += FF[i];
                FF[i] = 0;
            } else {
                while (o > 0 && FF[i]) {
                    ++ans;
                    o -= i;
                    --FF[i];
                }
                FF[i] = 0;
            }
        }

        return ans;
    }
};
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