1338. Reduce Array Size to The Half**

​​https://leetcode.com/problems/reduce-array-size-to-the-half/​​

题目描述

Given an array ​​arr​​. You can choose a set of integers and remove all the occurrences of these integers in the array.

Return the minimum size of the set so that at least half of the integers of the array are removed.

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​​

C++ 实现 1

保存每个元素出现的个数, 按照出现次数由高到低的顺序移除元素, 直到移除的元素个数至少是原数组大小的一半. 这个步骤既可以用最大堆实现, 也可以排序实现.

class Solution {
private:
    struct Comp {
        bool operator()(const pair<int, int> &p, const pair<int, int> &q) {
            return p.second < q.second;
        }
    };
public:
    int minSetSize(vector<int>& arr) {
        unordered_map<int, int> record;
        for (auto &a : arr) record[a] ++;
        priority_queue<pair<int, int>, vector<pair<int, int>>, Comp> q;
        for (auto &p : record) q.push(p);
        int res = 0, size = 0;
        while (!q.empty()) {
            auto p = q.top();
            q.pop();
            size += p.second;
            res ++;
            if (size >= arr.size() / 2) break;
        }
        return res;
    }
};