[LeetCode] 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

最大的团队表现值。

给定两个整数 n 和 k，以及两个长度为 n 的整数数组 speed 和 efficiency。现有 n 名工程师，编号从 1 到 n。其中 speed[i] 和 efficiency[i] 分别代表第 i 位工程师的速度和效率。

从这 n 名工程师中最多选择 k 名不同的工程师，使其组成的团队具有最大的团队表现值。

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」。

请你返回该团队的​​​​​​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。

思路是贪心，具体做法设计排序 + 最小堆。

注意题目对团队表现值的定义，表现值 = sum(speed of k engineers) * min(efficiency)。为了使最后的表现值更大，这里我们需要从 n 名工程师里挑出 k 名工程师，但是我们既要使他们的速度和越大，又要使这 k 名工程师的效率值的最小值越大，我们没法单纯对其中一个变量排序。这里我的思路是把工程师的速度 i 和效率 j 组合成一个二维数组 engineer[i][j]，然后对这个二维数组按照效率值降序排序，这样我每遇到一个工程师 engineer[i]，他一定是当前我已经遇到过的工程师里效率最低的那个。

遍历二维数组，将每个工程师的速度 i 加入最小堆，并把当前工程师的速度加到最后表现值的计算当中。当堆中元素个数 >= k 的时候，我们可以把速度最小的那个工程师弹出。具体细节参见代码。

时间O(nlogk)

空间O(k)

Java实现

class Solution {
    public int maxPerformance(int n, int[] speed, int[] efficiency, int k) {
        int MOD = (int) Math.pow(10, 9) + 7;
        // [efficiency, speed]
        int[][] engineers = new int[n][2];
        for (int i = 0; i < n; i++) {
            engineers[i][0] = efficiency[i];
            engineers[i][1] = speed[i];
        }
        // sort the efficiency in decreasing order
        Arrays.sort(engineers, (a, b) -> b[0] - a[0]);
        PriorityQueue<Integer> queue = new PriorityQueue<>();
        long tPerformance = 0;
        long tSpeed = 0;
        for (int i = 0; i < n; i++) {
            if (queue.size() >= k) {
                tSpeed -= queue.poll();
            }
            queue.offer(engineers[i][1]);
            tSpeed += engineers[i][1];
            tPerformance = Math.max(tPerformance, tSpeed * engineers[i][0]);
        }
        return (int) (tPerformance % MOD);
    }
}

 

LeetCode 题目总结