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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 i-th 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 <= 10^5speed.length == nefficiency.length == n1 <= speed[i] <= 10^51 <= efficiency[i] <= 10^81 <= 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
kengineers in the team, pop the one with the leastspeed, and deduct hisspeedfromsum. - try to update the answer using
sum * speed[i].
The idea behind
// OJ: https://leetcode.com/problems/maximum-performance-of-a-team/
// Time: O(NlogN)
// Space: O(N)
// Ref: https://leetcode.com/problems/maximum-performance-of-a-team/discuss/539687/JavaC%2B%2BPython-Priority-Queue
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);
}
};
-
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); } } ############ 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); } } -
// OJ: https://leetcode.com/problems/maximum-performance-of-a-team/ // Time: O(NlogN) // Space: O(N) // Ref: https://leetcode.com/problems/maximum-performance-of-a-team/discuss/539687/JavaC%2B%2BPython-Priority-Queue 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; } }; -
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 } func max(a, b int) int { if a > b { return a } return b } type hp struct{ sort.IntSlice } func (h *hp) Push(v interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) } func (h *hp) Pop() interface{} { 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] }