<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) {
        ios::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);

        int limit = arr.size(), count = 0, half = limit / 2, sum = 0;

#if 0
        //1.選擇m_map(這題值的range比較大)
        unordered_map<int,int> m_map;
        for(int num : arr) {
            ++m_map[num];
        }

        //2.把值放到priority_queue排序
        priority_queue<int> m_pq;
        for(auto& item : m_map) {
            m_pq.push(item.second);
        }
#else
        //1.選擇vector
        vector<int> counts(1e5 + 1);
        for(int num : arr) {
            ++counts[num];
        }

        //2.把值放到priority_queue排序
        priority_queue<int> m_pq;
        for(int count : counts) {
            if(count > 0)
                m_pq.push(count);
        }
#endif
        //3.把值pop出來直到累計總數已至少一半以上
        while(sum < half) {
            sum += m_pq.top();
            m_pq.pop();
            ++count;
        }
        return count;
    }
};
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