LeetCode解题报告（393) -- 1383. Maximum Performance of a Team

Problem

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:

1 2	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

Analysis

题目的通过两个数组，给出了每个工人的速度和效率，定义整个团队的performance是最小的效率乘以速度之和。要求我们从中选出k个成员，找到能够最大化的performance。首先，这里的关键应该是最小的效率，因为每个人的速度都要乘以它，所以我们优先考虑效率。我们先根据效率从大到小排序，然后把每一个成员的速度加起来，全局维护一个极大值。

然后，问题就转化为如果人数超过了k个人怎么办？因为效率是从大到小遍历的，所以当处理到第i个工人时，肯定是以第i个工人的效率为最低效率，关键是速度。因为我们要的是速度之和最大，很自然地我们就需要剔除掉速度最小的一个工人了，所以我们需要用一个最小堆去记录。当超过k个人之后，剔除掉速度最小的即可。

Solution

无。

Code

class Solution {
public:
    struct Engineer {
        int speed, efficiency;
        bool operator < (const Engineer& rhs) const {
            return speed > rhs.speed;
        }
    };
    
    int maxPerformance(int n, vector<int>& speed, vector<int>& efficiency, int k) {
        vector<Engineer> v;
        priority_queue<Engineer> q;
        for (int i = 0; i < n; i++) {
            v.push_back({speed[i], efficiency[i]});
        }
        
        sort(v.begin(), v.end(), [] (const Engineer &u, const Engineer &v) {
            return u.efficiency > v.efficiency;
        });
        
        long long result = 0, sum = 0;
        for (int i = 0; i < n; i++) {
            long long minEfficiency = v[i].efficiency;
            long long sumSpeed = sum + v[i].speed;
            result = max(result, sumSpeed * minEfficiency);
            q.push(v[i]);
            sum += v[i].speed;
            
            if (q.size() == k) {
                sum -= q.top().speed;
                q.pop();
            }
        }
        int MOD = 1e9 + 7;
        return result % MOD;
    }
};

Summary

这道题难度中等，面对这种有数量限制的题目，一般采用的是滑动窗口（连续）或者堆（非连续）。这道题目的分享到这里，感谢你的支持！