LeetCode 1383. Maximum Performance of a Team Solution in Java, C++, Python & Go | Explanation + Code

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 its engineers' speeds multiplied by the minimum efficiency among its 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 <= 105
• speed.length == n
• efficiency.length == n
• 1 <= speed[i] <= 105
• 1 <= efficiency[i] <= 108

Solutions

Solution 1

PythonJavaC++Go

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(code-box)

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); } }(code-box)

class Solution { public: int maxPerformance(int n, vector<int>& speed, vector<int>& efficiency, int k) { vector<pair<int, int>> t(n); for (int i = 0; i < n; ++i) t[i] = {-efficiency[i], speed[i]}; sort(t.begin(), t.end()); priority_queue<int, vector<int>, greater<int>> q; long long ans = 0, tot = 0; int mod = 1e9 + 7; for (auto& x : t) { int s = x.second, e = -x.first; tot += s; ans = max(ans, tot * e); q.push(s); if (q.size() == k) { tot -= q.top(); q.pop(); } } return (int) (ans % mod); } };(code-box)

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 } type hp struct{ sort.IntSlice } func (h *hp) Push(v any) { h.IntSlice = append(h.IntSlice, v.(int)) } func (h *hp) Pop() any { 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] }(code-box)