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 <= 10^5
  • speed.length == n
  • efficiency.length == n
  • 1 <= speed[i] <= 10^5
  • 1 <= efficiency[i] <= 10^8
  • 1 <= k <= n

Related Topics:
Greedy, Sort

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 the sum.
  • If after the addition there are more than k engineers in the team, pop the one with the least speed, and deduct his speed from sum.
  • 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);
    }
};

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);
    }
}

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