How do you solve Maximum Performance of a Team on LeetCode?

Maximum Performance of a Team (LeetCode #1383) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting engineers by efficiency helps in maximizing performance.

What is the brute force solution for Maximum Performance of a Team?

The brute force approach for Maximum Performance of a Team is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible combination of engineers to find the one that gives the maximum performance. This means checking all subsets of engineers, which can be computationally expensive. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Performance of a Team?

Maximum Performance of a Team uses the following concepts: Array, Greedy, Sorting, Heap (Priority Queue). The recommended patterns to study are: Greedy, Heap, Sorting.

What are the interview tips for Maximum Performance of a Team?

Think about how to optimize your solution before coding. Use data structures like heaps to manage dynamic data efficiently. Practice sorting and combining data in different ways.

What are common mistakes when solving Maximum Performance of a Team?

Not considering the minimum efficiency when calculating performance. Failing to manage the size of the team correctly with the heap.

#1383

Maximum Performance of a Team

Hard

ArrayGreedySortingHeap (Priority Queue)GreedyHeapSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

By sorting engineers based on efficiency and using a min-heap to keep track of the top speeds, we can efficiently calculate the maximum performance without checking all combinations.

⚙️

Algorithm

4 steps

1Step 1: Pair each engineer's speed with their efficiency and sort these pairs by efficiency in descending order.
2Step 2: Use a min-heap to keep track of the top k speeds as we iterate through the sorted list.
3Step 3: For each engineer, calculate the current performance using the minimum efficiency (current engineer's efficiency) and the sum of speeds in the heap.
4Step 4: Update the maximum performance found.

solution.py15 lines

1# Full working Python code
2import heapq
3
4def maxPerformance(n, speed, efficiency, k):
5    engineers = sorted(zip(efficiency, speed), reverse=True)
6    max_perf = 0
7    speed_sum = 0
8    min_heap = []
9    for eff, spd in engineers:
10        heapq.heappush(min_heap, spd)
11        speed_sum += spd
12        if len(min_heap) > k:
13            speed_sum -= heapq.heappop(min_heap)
14        max_perf = max(max_perf, speed_sum * eff)
15    return max_perf % (10**9 + 7)

ℹ

Complexity note: The sorting step takes O(n log n), and maintaining the min-heap takes O(n) in total for all engineers, leading to an overall complexity of O(n log n).

1Sorting engineers by efficiency helps in maximizing performance.
2Using a min-heap allows efficient management of the top speeds.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.