<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 = arr.size();
        int n = arr.size();
        map<int,int> mp;
        for(int i =0;i<arr.size();i++){
            mp[arr[i]]++;
        }

        vector<int> freq;
        for(auto it = mp.begin();it!=mp.end();it++){
            freq.push_back(it->second);
        }
        sort(freq.begin(),freq.end());

        int sum = n/2;
        int count = 0;
        int i = freq.size()-1;
        while(i>=0 && sum >0){
            sum = sum - freq[i--];
            count++;
        }

        return count;
        
    }
};
/**

[3,3,3,3,5,5,5,2,2,7]

2 - 2
3 - 4
5 - 3
7 - 1



**/
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