<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>

#include <cmath>
class Solution {
public:
    int minSetSize(vector<int>& arr) {
        unordered_map<int, int> m;
        int n = arr.size();
        for(const int& i : arr){
            m[i]++;
        }
        vector<int> freqs(n+1, 0);  // freqs[i] = number of elements repeated i times 
        for(auto i = m.begin(); i!=m.end(); ++i) {
            freqs[i->second]++;
        }
        int posn = n;
        int to_remove = ceil(n/2);
        int removed = 0;

        while(to_remove > 0) {
            if(freqs[posn] == 0) {
                posn--;
            } else {
                to_remove-=posn;
                freqs[posn]--;
                removed++;
            }
        }

        return removed;

    }
};
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