1338. Reduce Array Size to The Half

问题：

给出一个数组，求至少拿出几个元素，使得他们出现的次数之和，>=总元素个数的一半

Example 1:
Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: 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 size greater than half of the size of the old array.

Example 2:
Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.

Example 3:
Input: arr = [1,9]
Output: 1

Example 4:
Input: arr = [1000,1000,3,7]
Output: 1

Example 5:
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
 

Constraints:
1 <= arr.length <= 10^5
arr.length is even.
1 <= arr[i] <= 10^5

　　

解法：

使用unordered_map存储每个元素出现的次数，

为了排序出现次数，首先将次数放入一个vector数列coutorder中。

对coutorder进行降序排序，

从头累计，sum>=1/2 arr.size()

则返回，元素个数。

代码参考：

class Solution {
public:
    int minSetSize(vector<int>& arr) {
        int sum=0;
        int res=0;
        unordered_map<int, int>cout;
        vector<int>coutorder;
        for(int a:arr){
            cout[a]++;
        }
        for(auto c:cout){
            coutorder.push_back(c.second);
        }
        sort(coutorder.rbegin(), coutorder.rend());
        for(int s:coutorder){
            sum+=s;
            res++;
            if(sum>=arr.size()/2) return res;
        }
        return res;
    }
};