README_EN.md

1383. Maximum Performance of a Team

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Description

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

Solutions

Python3

class Solution:
    def maxPerformance(self, n: int, speed: List[int], efficiency: List[int], k: int) -> int:
        team = [(s, e) for s, e in zip(speed, efficiency)]
        team.sort(key=lambda x: -x[1])
        q = []
        t = 0
        ans = 0
        mod = int(1e9) + 7
        for s, e in team:
            t += s
            ans = max(ans, t * e)
            heappush(q, s)
            if len(q) >= k:
                t -= heappop(q)
        return ans % mod

Java

class Solution {
    public int maxPerformance(int n, int[] speed, int[] efficiency, int k) {
        int[][] team = new int[n][2];
        for (int i = 0; i < n; ++i) {
            team[i] = new int[]{speed[i], efficiency[i]};
        }
        Arrays.sort(team, (a, b) -> b[1] - a[1]);
        PriorityQueue<Integer> q = new PriorityQueue<>((a, b) -> a - b);
        long t = 0;
        long ans = 0;
        int mod = (int) 1e9 + 7;
        for (int[] x : team) {
            int s = x[0], e = x[1];
            t += s;
            ans = Math.max(ans, t * e);
            q.offer(s);
            if (q.size() >= k) {
                t -= q.poll();
            }
        }
        return (int) (ans % mod);
    }
}

C++

class Solution {
public:
    int maxPerformance(int n, vector<int>& speed, vector<int>& efficiency, int k) {
        vector<pair<int, int>> team;
        for (int i = 0; i < n; ++i) team.push_back({-efficiency[i], speed[i]});
        sort(team.begin(), team.end());
        priority_queue<int, vector<int>, greater<int>> q;
        long long ans = 0;
        int mod = 1e9 + 7;
        long long t = 0;
        for (auto& x : team)
        {
            int s = x.second, e = -x.first;
            t += s;
            ans = max(ans, e * t);
            q.push(s);
            if (q.size() >= k)
            {
                t -= q.top();
                q.pop();
            }
        }
        return (int) (ans % mod);
    }
};

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