<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>

#pragma GCC target("avx, mmx, sse2, sse3, sse4")

auto disableSync = [](void) noexcept -> int
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    return 0;
}();

class Solution final
{
public:
    int minSetSize(const std::vector<int>& nums) const noexcept
    {
        std::unordered_map<int, int> counter;
        for (const int num : nums)
            ++counter[num];
        std::vector<int> counts;
        counts.reserve(counter.size() + 4);
        for (const auto& [_, count] : counter)
            counts.push_back(count);
        std::sort(counts.rbegin(), counts.rend());
        int halfSize = nums.size() >> 1;
        for (auto it = counts.cbegin(); true; ++it)
            if ((halfSize -= *it) <= 0)
                return it - counts.begin() + 1;
        return INT_MAX; 
    }
};
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