1338. Reduce Array Size to The Half**

1338. Reduce Array Size to The Half**

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

題目描述

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++ 實現 1

保存每個元素出現的個數, 按照出現次數由高到低的順序移除元素, 直到移除的元素個數至少是原數組大小的一半. 這個步驟既可以用最大堆實現, 也可以排序實現.

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;
    }
};