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Maximum Performance of a Team in C++
Suppose there are n engineers. They are numbered from 1 to n and we also have two arrays: speed and efficiency, here speed[i] and efficiency[i] represent the speed and efficiency for the ith engineer. We have to find the maximum performance of a team composed of at most k engineers. The answer may be very large so return it modulo 10^9 + 7.
Here the performance of a team is the sum of their engineers' speeds multiplied by the minimum efficiency among their engineers.
So, if the input is like n = 6, speed = [1,5,8,2,10,3], efficiency = [9,7,2,5,4,3], k = 2, then the output will be 60, as we have the maximum performance of the team by selecting engineer with speed 10 and efficiency 4 and engineer with speed 5 and efficiency 7. That is, performance = (10 + 5) * min(4, 7) = 60.
To solve this, we will follow these steps −
ret := 0
Define one 2D array v
for initialize i := 0, when i < n, update (increase i by 1), do −
insert { e[i], s[i] } at the end of v
sort the array v in reverse order
Define one priority queue pq
sum := 0
for initialize i := 0, when i < n, update (increase i by 1), do −
if size of pq is same as k, then −
sum := top element of sum - pq
delete element from pq
sum := sum + v[i, 1]
insert v[i, 1] into pq
ret := maximum of ret and sum * v[i, 0]
return ret mod (1^9 + 7)
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int maxPerformance(int n, vector<int>& s, vector<int>& e, int k){
long long int ret = 0;
vector<vector<int> > v;
for (int i = 0; i < n; i++) {
v.push_back({ e[i], s[i] });
}
sort(v.rbegin(), v.rend());
priority_queue<int, vector<int>, greater<int> > pq;
long long int sum = 0;
for (int i = 0; i < n; i++) {
if (pq.size() == k) {
sum -= pq.top();
pq.pop();
}
sum += v[i][1];
pq.push(v[i][1]);
ret = max(ret, sum * v[i][0]);
}
return ret % (long long int)(1e9 + 7);
}
};
main(){
Solution ob;
vector<int> v = {1,5,8,2,10,3};
vector<int> v1 = {9,7,2,5,4,3};
cout << (ob.maxPerformance(6,v,v1,2));
}Input
6, {1,5,8,2,10,3}, {9,7,2,5,4,3}, 2Output
60
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