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 <= 105speed.length == nefficiency.length == n1 <= speed[i] <= 1051 <= efficiency[i] <= 108
Solutions
Solution 1
Intuition
Code
class Solution {
public:
int maxPerformance(int n, vector<int>& speed, vector<int>& efficiency, int k) {
int MOD = 1e9 + 7;
vector<pair<int, int>> engineers;
for (int i = 0; i < n; i++)
engineers.push_back({ speed[i], efficiency[i] });
// sort engineers by their efficiencies, from big to small
sort(engineers.begin(), engineers.end(), [&](auto& a, auto& b) {
return a.second > b.second;
});
// store their speed, the slowest is on top, could be pop
priority_queue<int, vector<int>, greater<int>> pq;
long speedSum = 0, res = 0;
for (auto& [s, e] : engineers) {
pq.push(s);
speedSum += s;
if (pq.size() > k) {
// pop the slowest guy
speedSum -= pq.top();
pq.pop();
}
// note because we sort engineers by their efficiencies, so the e is the smaller one
res = max(res, speedSum * e);
}
return res % MOD;
}
};
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