# LeetCode 1383. Maximum Performance of a Team

原题链接在这里：https://leetcode.com/problems/maximum-performance-of-a-team/description/

题目：

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`

题解：

The question is ask to maximize sum(speed) * min(efficiency).

Then try every efficiency from high -> low, and meanwhile maintain the as larget as possible speed group. Keep adding spped to the total speed.

If the group size == k, delete the smallest speed.

We could first sort array based on efficiency.

And use a minHeap to record the group speed to easily poll the smallest speed.

Time Complexity: O(nlogn).

Space: O(n).

AC Java:

1 class Solution { 2 public int maxPerformance(int n, int[] speed, int[] efficiency, int k) { 3 if(speed == null || speed.length == 0 || speed.length != efficiency.length || k <= 0){ 4 return 0; 5 } 6 7 int mod = (int)1e9 + 7; 8 int[][] arr = new int[n][2]; 9 for(int i = 0; i < n; i++){ 10 arr[i] = new int[]{efficiency[i], speed[i]}; 11 } 12 13 Arrays.sort(arr, (a, b) -> b[0] - a[0]); 14 PriorityQueue<Integer> minHeap = new PriorityQueue<>(); 15 long res = 0; 16 long sum = 0; 17 for(int [] a : arr){ 18 if(minHeap.size() == k){ 19 sum -= minHeap.poll(); 20 } 21 22 sum += a[1]; 23 minHeap.add(a[1]); 24 res = Math.max(res, sum * a[0]); 25 } 26 27 return (int)(res % mod); 28 } 29 }