原文链接: https://leetcode-cn.com/problems/maximum-performance-of-a-team

英文原文

You are given two integers n and k and two integer arrays speed and efficiency both of length n. There are n engineers numbered from 1 to n. speed[i] and efficiency[i] represent the speed and efficiency of the ith engineer respectively.

Choose at most k different engineers out of the n engineers to form a team with the maximum performance.

The performance of a team is the sum of their engineers' speeds multiplied by the minimum efficiency among their engineers.

Return the maximum performance of this team. Since the answer can be a huge number, return it modulo 109 + 7.

 

Example 1:

Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2
Output: 60
Explanation: 
We have the maximum performance of the team by selecting engineer 2 (with speed=10 and efficiency=4) and engineer 5 (with speed=5 and efficiency=7). That is, performance = (10 + 5) * min(4, 7) = 60.

Example 2:

Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3
Output: 68
Explanation:
This is the same example as the first but k = 3. We can select engineer 1, engineer 2 and engineer 5 to get the maximum performance of the team. That is, performance = (2 + 10 + 5) * min(5, 4, 7) = 68.

Example 3:

Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4
Output: 72

 

Constraints:

• 1 <= k <= n <= 105
• speed.length == n
• efficiency.length == n
• 1 <= speed[i] <= 105
• 1 <= efficiency[i] <= 108

中文题目

公司有编号为 1 到 n 的 n 个工程师，给你两个数组 speed 和 efficiency ，其中 speed[i] 和 efficiency[i] 分别代表第 i 位工程师的速度和效率。请你返回由最多 k 个工程师组成的 ​​​​​​最大团队表现值 ，由于答案可能很大，请你返回结果对 10^9 + 7 取余后的结果。

团队表现值 的定义为：一个团队中「所有工程师速度的和」乘以他们「效率值中的最小值」。

 

示例 1：

输入：n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2
输出：60
解释：
我们选择工程师 2（speed=10 且 efficiency=4）和工程师 5（speed=5 且 efficiency=7）。他们的团队表现值为 performance = (10 + 5) * min(4, 7) = 60 。

示例 2：

输入：n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3
输出：68
解释：
此示例与第一个示例相同，除了 k = 3 。我们可以选择工程师 1 ，工程师 2 和工程师 5 得到最大的团队表现值。表现值为 performance = (2 + 10 + 5) * min(5, 4, 7) = 68 。

示例 3：

输入：n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4
输出：72

 

提示：

• 1 <= n <= 10^5
• speed.length == n
• efficiency.length == n
• 1 <= speed[i] <= 10^5
• 1 <= efficiency[i] <= 10^8
• 1 <= k <= n

通过代码

高赞题解

思路

直观的解法是按照效率进行降序排序，每个人作为最低效率时，在其左侧找出至多K - 1个最大速度即可(再加上这个人的速度组成K个)，这一过程可以用堆，时间复杂度O(nlg(k-1))

代码

class Solution {
    public int maxPerformance(int n, int[] speed, int[] efficiency, int k) {
        int[][] items = new int[n][2];
        for(int i = 0 ; i < n ; i++){
            items[i][0] = speed[i];
            items[i][1] = efficiency[i];
        }
        Arrays.sort(items, new Comparator<int[]>(){
           @Override
            public int compare(int[] a, int[] b){
                return b[1] - a[1];
            }
        });
        PriorityQueue<Integer> queue = new PriorityQueue<>();
        long res = 0, sum = 0;
        for(int i = 0 ; i < n ; i++){
            if(queue.size() > k - 1){
                sum -= queue.poll();
            }
            res = Math.max(res, (sum + items[i][0])* items[i][1]);
            queue.add(items[i][0]);
            sum += items[i][0];
        }
        return (int)(res % ((int)1e9 + 7));
    }
}

统计信息

通过次数	提交次数	AC比率
4696	14430	32.5%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言