Maximum Performance of a Team #1 Most Asked

Hard

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Topics:

ArraysGreedy Algorithms

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 i^th 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

Try coding it in LeetCode

Maximum Performance of a Team #1 Most Asked

Hard

Citrix

1 view

Topics:

ArraysGreedy Algorithms

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 i^th 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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the speed and efficiency values, and what data types will they be (e.g., integers, floats)?
Can speed or efficiency values be zero or negative?
What should I return if the input arrays are empty or null?
If multiple teams have the same maximum performance, is any one of them acceptable, or is there a specific criterion for choosing one?
Is `k` guaranteed to be less than or equal to the length of the input arrays?

Brute Force Solution

Approach

The goal is to find the best team by considering every possible team combination. We'll examine each potential team and calculate its performance, keeping track of the best one found so far. The brute force strategy looks at every possibility, even if it takes a long time.

Here's how the algorithm would work step-by-step:

Start by considering a team of just one engineer. Calculate their 'performance score'.
Then, consider all possible teams of two engineers. For each team, calculate their 'performance score'.
Next, consider all possible teams of three engineers, and so on, up to the maximum allowed team size. For each team, calculate their 'performance score'.
For each team you consider, determine the overall performance using these two factors: sum up the 'speed' of all team members and find the 'efficiency' of the slowest team member.
Multiply the sum of the speeds by the efficiency of the slowest member. This gives you a 'performance score' for the team.
Keep track of the highest 'performance score' you have found so far.
After you have examined all possible team combinations, the highest 'performance score' you tracked is the answer.

Code Implementation

def max_performance_team_brute_force(number_of_engineers, engineer_speeds, engineer_efficiencies, max_team_size):
    maximum_performance = 0
    
    for team_size in range(1, min(number_of_engineers + 1, max_team_size + 1)):
        # Iterate through all possible combinations of engineers for current team size
        for combination in get_combinations(number_of_engineers, team_size):
            team_speed_sum = 0
            minimum_efficiency = float('inf')
            
            for engineer_index in combination:
                team_speed_sum += engineer_speeds[engineer_index]
                minimum_efficiency = min(minimum_efficiency, engineer_efficiencies[engineer_index])
            
            # Calculate the performance score for the current team
            team_performance = team_speed_sum * minimum_efficiency
            
            # Update the maximum performance if the current team's performance is better
            maximum_performance = max(maximum_performance, team_performance)

    return int(maximum_performance % (10**9 + 7))

def get_combinations(number_of_engineers, team_size):
    combinations = []
    
    def backtrack(start, current_combination):
        if len(current_combination) == team_size:
            combinations.append(current_combination[:])
            return
        
        # Ensures that we do not go out of the index bounds
        for i in range(start, number_of_engineers):
            current_combination.append(i)
            backtrack(i + 1, current_combination)
            current_combination.pop()
            
    backtrack(0, [])
    return combinations

Big(O) Analysis

Time Complexity

O(n^k) – The algorithm considers all possible teams from size 1 up to k (maximum allowed team size) from a pool of n engineers. Generating all teams of size i from n engineers has a complexity of n choose i, which is O(n^i). Because we generate teams from size 1 to k, we sum the cost of each team size: O(n^1) + O(n^2) + ... + O(n^k). The largest term, O(n^k), dominates the overall complexity, resulting in O(n^k).

Space Complexity

O(1) – The algorithm iterates through all possible team combinations and calculates the performance score for each team. While it keeps track of the highest performance score found so far, this requires only a single variable. No auxiliary data structures that scale with the input size (N, representing the number of engineers) are used. Therefore, the auxiliary space complexity is constant.

Optimal Solution

Approach

To maximize the team's performance, we need to strategically select engineers based on their speed and efficiency. We prioritize faster engineers but must also consider the impact of each engineer's specialization on the team's overall effectiveness. In the end, we'll have the best possible group for the job.

Here's how the algorithm would work step-by-step:

First, we sort all engineers from fastest to slowest in terms of efficiency.
Then, we consider each engineer in order from fastest to slowest, potentially including them in our team.
As we build the team, we track the total efficiency of the team and identify the minimum specialization rating of the team.
We compute the team's performance which equals the total efficiency multiplied by the minimum specialization rating.
The goal is to try different team compositions to see which one yields the highest possible team performance.
If at any point the number of team members exceeds what is allowed, we remove slower members of the team.
Keep track of the largest performance score calculated during the process. This value is the result.

Code Implementation

import heapq

def max_performance(number_of_engineers, engineer_speeds, engineer_efficiencies, team_size):

    engineers = sorted(zip(engineer_speeds, engineer_efficiencies), key=lambda x: x[1], reverse=True)
    team_performance = 0
    team_speed_total = 0
    min_efficiency = float('inf')
    speed_heap = []

    for engineer_speed, engineer_efficiency in engineers:
        # Add the current engineer's speed to the total.
        team_speed_total += engineer_speed
        heapq.heappush(speed_heap, engineer_speed)

        # Limit the team size if it exceeds the maximum allowed.
        if len(speed_heap) > team_size:
            team_speed_total -= heapq.heappop(speed_heap)

        # Calculate team performance.
        team_performance = max(team_performance, team_speed_total * engineer_efficiency)

    return team_performance % (10**9 + 7)

Big(O) Analysis

Time Complexity

O(n log n) – The dominant operation is sorting the engineers based on efficiency, which takes O(n log n) time. We then iterate through the sorted engineers once (O(n)). Inside the loop, we maintain a priority queue (or similar data structure) of limited size k (where k is the maximum team size). Adding and removing elements from this priority queue takes O(log k) time, and since k is bounded, we can consider it O(log n) in the worst case. The overall time complexity is therefore O(n log n) + O(n log n), which simplifies to O(n log n).

Space Complexity

O(N) – The algorithm sorts the engineers based on efficiency, potentially requiring an auxiliary array of size N to store the sorted indices, where N is the number of engineers. Additionally, a priority queue (or similar data structure) is used to maintain the speeds of the selected engineers. In the worst case, this queue could hold up to K engineers, although K is bounded by N. Furthermore, variables to store running totals and the result use constant space. Thus, the dominant space complexity is determined by the sorting and the priority queue, both of which could potentially use O(N) space in the worst case.

Edge Cases

Case	How to Handle
Empty speed or efficiency array	Return 0 immediately as no team can be formed.
Null speed or efficiency array	Throw an IllegalArgumentException or similar error.
Speed and efficiency arrays have different lengths	Throw an IllegalArgumentException indicating arrays must have equal length.
k is zero	Return 0 immediately, as no team can be formed.
k is greater than the array length	Use the array length as the effective k since we can only choose up to all engineers.
Large speed values leading to integer overflow	Use long data type for intermediate calculations to prevent overflow before applying the modulo.
All efficiencies are the same	The algorithm should still work correctly, selecting the k fastest engineers.
Speed or efficiency values are negative	Throw an IllegalArgumentException indicating speeds and efficiencies must be non-negative.

Empty speed or efficiency array

How to Handle:

Return 0 immediately as no team can be formed.

Null speed or efficiency array

How to Handle:

Throw an IllegalArgumentException or similar error.

Speed and efficiency arrays have different lengths

How to Handle:

Throw an IllegalArgumentException indicating arrays must have equal length.

k is zero

How to Handle:

Return 0 immediately, as no team can be formed.

k is greater than the array length

How to Handle:

Use the array length as the effective k since we can only choose up to all engineers.

Large speed values leading to integer overflow

How to Handle:

Use long data type for intermediate calculations to prevent overflow before applying the modulo.

All efficiencies are the same

How to Handle:

The algorithm should still work correctly, selecting the k fastest engineers.

Speed or efficiency values are negative

How to Handle:

Throw an IllegalArgumentException indicating speeds and efficiencies must be non-negative.