1383. Maximum Performance of a Team

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 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

• class Solution {
private static final int MOD = (int) 1e9 + 7;

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

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

• class Solution:
def maxPerformance(
self, n: int, speed: List[int], efficiency: List[int], k: int
) -> int:
t = sorted(zip(speed, efficiency), key=lambda x: -x[1])
ans = tot = 0
mod = 10**9 + 7
h = []
for s, e in t:
tot += s
ans = max(ans, tot * e)
heappush(h, s)
if len(h) == k:
tot -= heappop(h)
return ans % mod


• func maxPerformance(n int, speed []int, efficiency []int, k int) int {
t := make([][]int, n)
for i, s := range speed {
t[i] = []int{s, efficiency[i]}
}
sort.Slice(t, func(i, j int) bool { return t[i][1] > t[j][1] })
var mod int = 1e9 + 7
ans, tot := 0, 0
pq := hp{}
for _, x := range t {
s, e := x[0], x[1]
tot += s
ans = max(ans, tot*e)
heap.Push(&pq, s)
if pq.Len() == k {
tot -= heap.Pop(&pq).(int)
}
}
return ans % mod
}

type hp struct{ sort.IntSlice }

func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
a := h.IntSlice
v := a[len(a)-1]
h.IntSlice = a[:len(a)-1]
return v
}
func (h *hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }