<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) {
        array<int, 100001> freq{0};
        auto [mn_it, mx_it] = minmax_element(arr.begin(), arr.end());
        for (auto num : arr) {
            ++freq[num];
        }
        vector<int> amount;
        for (int num = *mn_it; num <= *mx_it; ++num) {
            if (freq[num] > 0) {
                amount.push_back(freq[num]);
            }
        }
        sort(amount.begin(), amount.end(), greater<int>());
        int sum{0}, ans{0}, half = (arr.size() + 1) / 2;
        for (int count : amount) {
            sum += count;
            ++ans;
            if (sum >= half) {
                return ans;
            }
        }
        return ans;
    }
};
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