What is the time complexity of Maximum Performance of a Team?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Maximum Performance of a Team?

This problem uses Array, Greedy, Sorting, Heap (Priority Queue). Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Performance of a Team?

Insight 1: Sorting engineers based on efficiency is crucial. By iterating through engineers in decreasing order of efficiency, we can maintain a running sum of speeds and use the current engineer's efficiency as the minimum efficiency for the team formed so far. Insight 2: A min-heap (priority queue) is essential for keeping track of the 'k' engineers with the highest speeds at any given point. This allows us to efficiently remove the engineer with the lowest speed when the team size exceeds 'k'.

Which companies ask LeetCode 1383. Maximum Performance of a Team in interviews?

This problem is frequently asked at Citrix, DE Shaw, DoorDash, Flipkart, PhonePe.

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Maximum Performance of a Team - LeetCode 1383 Solution

Maximum Performance of a Team - Complete Solution Guide

Maximum Performance of a Team is LeetCode problem 1383, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem requires finding the maximum performance of a team formed by selecting at most 'k' engineers from a pool of 'n' engineers. Each engineer has a 'speed' and 'efficiency'. The performance of a team is calculated as the sum of the speeds of the selected engineers multiplied by the minimum efficiency among those engineers. The goal is to maximize this performance, returning the result modulo 10^9 + 7. The inputs are two arrays 'speed' and 'efficiency' of length 'n', an integer 'n' representing the number of engineers, and an integer 'k' representing the maximum number of engineers to select. The output is the maximum possible performance of a team of at most 'k' engineers, modulo 10^9 + 7.

Solution Approach

The solution uses a greedy approach. First, the engineers are sorted in descending order of their efficiency. Then, we iterate through the sorted engineers. For each engineer, we add their speed to a min-heap and update the total speed sum. If the number of engineers in the heap exceeds 'k', we remove the engineer with the smallest speed from the heap and subtract their speed from the total speed sum. Finally, we calculate the current team's performance by multiplying the total speed sum by the current engineer's efficiency (which is the minimum efficiency of the team formed so far) and update the maximum performance if necessary. The modulo operation is applied before returning the result.

Step-by-Step Algorithm

Step 1: Create pairs of (efficiency, speed) for each engineer.
Step 2: Sort the engineers in descending order of efficiency.
Step 3: Initialize a min-heap to store the speeds of the selected engineers.
Step 4: Initialize a variable to track the sum of the speeds of the selected engineers.
Step 5: Initialize a variable to store the maximum performance.
Step 6: Iterate through the sorted engineers:
a. Add the current engineer's speed to the min-heap.
b. Add the current engineer's speed to the total speed sum.
c. If the number of engineers in the heap is greater than 'k':
i. Remove the engineer with the smallest speed from the heap.
ii. Subtract the removed engineer's speed from the total speed sum.
d. Calculate the current team's performance as (total speed sum * current engineer's efficiency).
e. Update the maximum performance if the current performance is greater.
Step 7: Return the maximum performance modulo (10^9 + 7).

Key Insights

Insight 1: Sorting engineers based on efficiency is crucial. By iterating through engineers in decreasing order of efficiency, we can maintain a running sum of speeds and use the current engineer's efficiency as the minimum efficiency for the team formed so far.
Insight 2: A min-heap (priority queue) is essential for keeping track of the 'k' engineers with the highest speeds at any given point. This allows us to efficiently remove the engineer with the lowest speed when the team size exceeds 'k'.
Insight 3: Since the performance can be very large, we need to apply the modulo operator (10^9 + 7) at the end to prevent integer overflow.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Greedy, Sorting, Heap (Priority Queue).

Companies

Asked at: Citrix, DE Shaw, DoorDash, Flipkart, PhonePe.

Maximum Performance of a Team | LeetCode 1383

Hard

Array Greedy Sorting Heap (Priority Queue)

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.PriorityQueue;

class Solution {
    public int maxPerformance(int n, int[] speed, int[] efficiency, int k) {
        int[][] engineers = new int[n][2];
        for (int i = 0; i < n; i++) {
            engineers[i][0] = efficiency[i];
            engineers[i][1] = speed[i];
        }

        Arrays.sort(engineers, (a, b) -> b[0] - a[0]);

        PriorityQueue<Integer> speedHeap = new PriorityQueue<>();
        long currentSpeedSum = 0;
        long maxPerformance = 0;

        for (int[] engineer : engineers) {
            int e = engineer[0];
            int s = engineer[1];
            speedHeap.offer(s);
            currentSpeedSum += s;

            if (speedHeap.size() > k) {
                currentSpeedSum -= speedHeap.poll();
            }

            maxPerformance = Math.max(maxPerformance, currentSpeedSum * e);
        }

        return (int) (maxPerformance % (1000000000 + 7));
    }
}

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Problem Info

Description

Problem 1383: Maximum Performance of a Team - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting Heap (Priority Queue)

Companies

CitrixDE ShawDoorDashFlipkartPhonePe

Stats

Difficulty:Hard

Languages:4

Companies:5

Problem #:1383

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

14 steps

1
Step 1: Create pairs of (efficiency, speed) for each engineer.
2
Step 2: Sort the engineers in descending order of efficiency.
3
Step 3: Initialize a min-heap to store the speeds of the selected engineers.
4
Step 4: Initialize a variable to track the sum of the speeds of the selected engineers.
5
Step 5: Initialize a variable to store the maximum performance.
6
Step 6: Iterate through the sorted engineers:
7
a. Add the current engineer's speed to the min-heap.
8
b. Add the current engineer's speed to the total speed sum.
9
c. If the number of engineers in the heap is greater than 'k':
10
i. Remove the engineer with the smallest speed from the heap.
11
ii. Subtract the removed engineer's speed from the total speed sum.
12
d. Calculate the current team's performance as (total speed sum * current engineer's efficiency).
13
e. Update the maximum performance if the current performance is greater.
14
Step 7: Return the maximum performance modulo (10^9 + 7).

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the sorting step which uses a comparison based sorting algorithm. Heap operations are performed in a loop, but since k <= n, the heap operations contribute a factor no greater than log(n) to the overall time complexity, not changing the overall O(n log n) runtime. The space complexity depends on how the initial engineers are stored (O(n)), and the heap (O(k)).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Try all possible combinations of up to k engineers and compute performance for each combination. Find the maximum.

Time: O(nCk * k), where nCk is the binomial coefficient 'n choose k', which can be approximated as O(n^k) in the worst case.Space: O(k)

Trade-offs: This approach is simple to understand but extremely inefficient and will result in a Time Limit Exceeded (TLE) error for large input sizes. It's practical only for very small 'n' and 'k'.

✨

Key Insights

1Insight 1: Sorting engineers based on efficiency is crucial. By iterating through engineers in decreasing order of efficiency, we can maintain a running sum of speeds and use the current engineer's efficiency as the minimum efficiency for the team formed so far.
2Insight 2: A min-heap (priority queue) is essential for keeping track of the 'k' engineers with the highest speeds at any given point. This allows us to efficiently remove the engineer with the lowest speed when the team size exceeds 'k'.
3Insight 3: Since the performance can be very large, we need to apply the modulo operator (10^9 + 7) at the end to prevent integer overflow.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: When k = 1, the solution will select the engineer with the highest efficiency and calculate the performance based on their speed and efficiency.
Edge case 2: When k = n, the solution will consider all engineers, and the performance will be the sum of all speeds multiplied by the minimum efficiency among all engineers.
Edge case 3: When n is very large, the speed sum can overflow if using int. The C and C++ solutions use `long long` to prevent integer overflow.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the engineers by efficiency first. This is crucial for the greedy approach to work correctly.

✗Common mistake 2: Using the wrong data structure for the heap. A min-heap is needed to efficiently remove the engineer with the smallest speed when the team size exceeds 'k'.

✗Common mistake 3: Integer overflow. The speed sum and the final performance can be very large, so use `long` or `long long` to store these values and apply the modulo operator at the end to prevent overflow. Failing to do this can lead to incorrect results.

✗Common mistake 4: Forgetting the modulo operator at the end after finding the performance, resulting in a wrong answer due to large integer values.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource allocation. Imagine you have a set of tasks with associated cost and priority. You want to select a subset of tasks to maximize the total priority, while ensuring the total cost stays within a certain budget. The 'efficiency' can represent priority, and the 'speed' can represent the inverse of the cost.

🏦Real-world scenario 2: Portfolio optimization. In finance, you can think of 'speed' as the return of an investment, and 'efficiency' as the risk associated with the investment. The problem translates to selecting a portfolio of at most k investments to maximize the return, considering the minimum risk tolerance.

🎮Real-world scenario 3: Team formation for projects: In project management, 'efficiency' might represent the skill level of a team member, and 'speed' might represent their productivity. The task is to form a team with at most 'k' members to maximize project performance, where performance is determined by the combined productivity and the minimum skill level among the team members.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 2, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Performance of a Team"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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