Formatted question description: https://leetcode.ca/all/1383.html
1383. Maximum Performance of a Team (Hard)
There are n
engineers numbered from 1 to n
and two arrays: speed
and efficiency
, where speed[i]
and efficiency[i]
represent the speed and efficiency for the ith engineer respectively. Return the maximum performance of a team composed of at most k
engineers, since the answer can be a huge number, return this modulo 10^9 + 7.
The performance of a team is the sum of their engineers' speeds multiplied by the minimum efficiency among their engineers.
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 <= n <= 10^5
speed.length == n
efficiency.length == n
1 <= speed[i] <= 10^5
1 <= efficiency[i] <= 10^8
1 <= k <= n
Solution 1. Greedy
Intuition
For a given efficiency, we pick all works with the same or better efficiency. If the number of workers is greater than k
, we pick k
fastest workers.
Algorithm
We greedily try each engineer from the most efficient one to the least one.
For each engineer:
 first try adding him to the team and add his
speed[i]
to thesum
.  If after the addition there are more than
k
engineers in the team, pop the one with the leastspeed
, and deduct hisspeed
fromsum
.  try to update the answer using
sum * speed[i]
.
The idea behind
// OJ: https://leetcode.com/problems/maximumperformanceofateam/
// Time: O(NlogN)
// Space: O(N)
// Ref: https://leetcode.com/problems/maximumperformanceofateam/discuss/539687/JavaC%2B%2BPythonPriorityQueue
class Solution {
public:
int maxPerformance(int N, vector<int>& S, vector<int>& E, int K) {
vector<pair<int, int>> ps(N);
for (int i = 0; i < N; ++i) ps[i] = {E[i], S[i]};
sort(ps.begin(), ps.end());
long sum = 0, ans = 0;
priority_queue<int, vector<int>, greater<int>> pq;
for (int i = N  1; i >= 0; i) {
pq.push(ps[i].second);
sum += ps[i].second;
if (pq.size() > K) {
sum = pq.top();
pq.pop();
}
ans = max(ans, sum * ps[i].first);
}
return ans % (int)(1e9+7);
}
};
Java

class Solution { public int maxPerformance(int n, int[] speed, int[] efficiency, int k) { final int MODULO = 1000000007; int[][] speedEfficiencyArray = new int[n][2]; for (int i = 0; i < n; i++) { speedEfficiencyArray[i][0] = speed[i]; speedEfficiencyArray[i][1] = efficiency[i]; } Arrays.sort(speedEfficiencyArray, new Comparator<int[]>() { public int compare(int[] speedEfficiency1, int[] speedEfficiency2) { if (speedEfficiency1[1] != speedEfficiency2[1]) return speedEfficiency2[1]  speedEfficiency1[1]; else return speedEfficiency2[0]  speedEfficiency1[0]; } }); long maxPerformance = 0; PriorityQueue<Integer> priorityQueue = new PriorityQueue<Integer>(); long speedSum = 0; int minEfficiency = Integer.MAX_VALUE; for (int i = 0; i < k; i++) { int[] speedEfficiency = speedEfficiencyArray[i]; int curSpeed = speedEfficiency[0]; int curEfficiency = speedEfficiency[1]; priorityQueue.offer(curSpeed); speedSum += curSpeed; minEfficiency = Math.min(minEfficiency, curEfficiency); long curPerformance = speedSum * (long) minEfficiency; maxPerformance = Math.max(maxPerformance, curPerformance); } for (int i = k; i < n; i++) { int prevSpeed = priorityQueue.poll(); speedSum = prevSpeed; int[] speedEfficiency = speedEfficiencyArray[i]; int curSpeed = speedEfficiency[0]; int curEfficiency = speedEfficiency[1]; priorityQueue.offer(curSpeed); speedSum += curSpeed; minEfficiency = Math.min(minEfficiency, curEfficiency); long curPerformance = speedSum * (long) minEfficiency; maxPerformance = Math.max(maxPerformance, curPerformance); } return (int) (maxPerformance % MODULO); } }

// OJ: https://leetcode.com/problems/maximumperformanceofateam/ // Time: O(NlogN) // Space: O(N) // Ref: https://leetcode.com/problems/maximumperformanceofateam/discuss/539687/JavaC%2B%2BPythonPriorityQueue class Solution { public: int maxPerformance(int N, vector<int>& S, vector<int>& E, int K) { vector<pair<int, int>> ps(N); for (int i = 0; i < N; ++i) ps[i] = {E[i], S[i]}; sort(ps.begin(), ps.end()); long sum = 0, ans = 0, mod = 1e9 + 7; priority_queue<int, vector<int>, greater<int>> pq; for (int i = N  1; i >= 0; i) { pq.push(ps[i].second); sum += ps[i].second; if (pq.size() > K) { sum = pq.top(); pq.pop(); } ans = max(ans, sum * ps[i].first); } return ans % mod; } };

print("Todo!")