Most Profit Assigning Work

class Solution {
public:
    int calculateMaximumProfit(vector<int>& difficulty, vector<int>& profit, vector<int>& worker) {
        int highestSkill = *max_element(worker.begin(), worker.end());
        vector<int> jobProfit(highestSkill + 1, 0);
        for (int i = 0; i < difficulty.size(); i++) {
            if (difficulty[i] <= highestSkill)
                jobProfit[difficulty[i]] = max(jobProfit[difficulty[i]], profit[i]);
        }
        for (int i = 1; i <= highestSkill; i++)
            jobProfit[i] = max(jobProfit[i], jobProfit[i - 1]);
    
        int totalProfit = 0;
        for (auto skill : worker) {
            int profitAtSkill = jobProfit[skill];
            totalProfit += profitAtSkill;
        }
        return totalProfit;
    }
};

This C++ solution tackles the problem of determining the maximum profit from assigning jobs based on their difficulty, profit, and the skills of the workers. The calculateMaximumProfit function starts by identifying the maximum skill level among the workers using max_element. It initializes a jobProfit vector the size of the highest skill plus one, with all values set to zero.

Here's the approach taken:

• Iterate through each job:
  • If the job's difficulty is less than or equal to the highest skill, update the corresponding index in the jobProfit vector with the maximum profit achievable for that difficulty.
• Modify the jobProfit vector so that jobProfit[i] always represents the maximum profit that can be obtained with skill level i or less. This handles cases where there might be several jobs of varying difficulties but identical skill levels.
• Calculate the total profit by iterating through each worker's skill and summing up the maximum profit obtainable for each skill level from the previously populated jobProfit vector.

This solution ensures that each worker is assigned the job with the highest profit possible according to their skill sets without exceeding the worker's level of difficulty, leading to an optimization of the total profit from all assignments.

class Solution {
    
    public int findMaxProfit(
        int[] difficulty,
        int[] profit,
        int[] worker
    ) {
        int highestSkill = Arrays.stream(worker).max().getAsInt();
        int[] taskProfits = new int[highestSkill + 1];
    
        for (int i = 0; i < difficulty.length; i++) {
            if (difficulty[i] <= highestSkill) {
                taskProfits[difficulty[i]] = Math.max(taskProfits[difficulty[i]], profit[i]);
            }
        }
    
        for (int i = 1; i <= highestSkill; i++) {
            taskProfits[i] = Math.max(taskProfits[i], taskProfits[i - 1]);
        }
    
        int totalProfit = 0;
        for (int skill : worker) {
            totalProfit += taskProfits[skill];
        }
        return totalProfit;
    }
}

This Java solution tackles the problem of determining the most profit from assigning work based on differing worker skills and task difficulties. Take the following steps to understand the code's approach:

1. Calculate the maximum skill level among all workers. This helps in setting the size for the taskProfits array that will store the highest possible profit for each skill level up to the highest worker's skill.

2. Initialize an array, taskProfits, to record the best profit obtainable for each difficulty level up to the maximum skill level found.

3. Loop through each task defined by arrays difficulty and profit. For tasks within the skill range of at least one worker, update the corresponding index in the taskProfits array with the maximum profit for that difficulty level.

4. Traverse the taskProfits array from beginning to end. Ensure that each position holds the maximum profit up to that difficulty level, enabling workers with higher skill levels to achieve more profit from simpler tasks if those simpler tasks offer higher compensation.

5. Finally, sum up the profits for each worker based on their skill levels using the taskProfits information. Workers will earn the best profit for their skill level, irrespective of the individual task difficulties they can handle.

This approach efficiently calculates total profits using dynamic programming concepts, ensuring optimal profit assignment for a given set of workers and tasks.

class Solution:
    def calculateMaxProfit(
        self, skillLevels: List[int], earnings: List[int], employees: List[int]
    ) -> int:
        maxSkill = max(employees)
        taskDifficulty = [0] * (maxSkill + 1)
    
        for index in range(len(skillLevels)):
            if skillLevels[index] <= maxSkill:
                taskDifficulty[skillLevels[index]] = max(taskDifficulty[skillLevels[index]], earnings[index])
    
        for j in range(1, maxSkill + 1):
            taskDifficulty[j] = max(taskDifficulty[j], taskDifficulty[j - 1])
    
        totalProfit = 0
        for skill in employees:
            totalProfit += taskDifficulty[skill]
    
        return totalProfit

This Python solution tackles the problem of maximizing profit from assigning work based on skill levels and corresponding earnings. The function calculateMaxProfit takes three parameters: skillLevels, earnings, and employees. Here's a breakdown of how the code operates:

• Calculate the highest skill level among the employees to ensure that task assignments cater to the highest possible skill level present.
• Initiate a list taskDifficulty where its indices represent skill levels and values represent the maximum earning potential for that skill level.
• Populate the taskDifficulty list such that each skill level has the maximum earnings it can yield without exceeding the specific skill. This is achieved by iterating through the given skillLevels and earnings.
• Iterate from the lowest to the highest skill levels to fill out any gaps in taskDifficulty. This ensures that for any skill level without direct corresponding earnings, the best possible undervalued task is assigned.
• Initialize totalProfit. Iterate over each employee's skill level to accumulate potential earnings based on the populated taskDifficulty.
• Finally, return totalProfit, representing the maximum profit derived from optimally assigned tasks based on employee skills.

This approach efficiently maps tasks to skills, ensuring that every employee's skill is utilized to its utmost profit potential while preserving the complexity of O(N) with respect to the number of employees and their skill distribution.