README.md

1383. Maximum Performance of a Team

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 its engineers' speeds multiplied by the minimum efficiency among its 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 (Rust)

1. Solution

use std::collections::BinaryHeap;

impl Solution {
    pub fn max_performance(n: i32, speed: Vec<i32>, efficiency: Vec<i32>, k: i32) -> i32 {
        let mut engineers = efficiency.iter().zip(speed.iter()).collect::<Vec<_>>();
        let mut heap = BinaryHeap::new();
        let mut speed_sum = 0;
        let mut ret = 0;

        engineers.sort_unstable();

        for &(&e, &s) in engineers.iter().rev() {
            if heap.len() == k as usize {
                speed_sum += heap.pop().unwrap();
            }
            speed_sum += s as i64;
            heap.push(-s as i64);
            ret = ret.max(speed_sum * e as i64);
        }

        (ret % 1_000_000_007) as i32
    }
}