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LeetCode 3476: Maximize Profit from Task Assignment Solution

Master LeetCode problem 3476 (Maximize Profit from Task Assignment), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3476. Maximize Profit from Task Assignment

Medium
ArrayGreedySortingHeap (Priority Queue)

Problem Explanation

Explanation:

To maximize the profit from task assignment, we can first sort the tasks and then assign the tasks to the workers with the highest and lowest efficiency alternatively. This way, we can ensure that the total profit is maximized by choosing the most efficient worker for each task.

  1. Sort the tasks based on their difficulty.
  2. Sort the workers based on their efficiency.
  3. Iterate over the sorted tasks and assign each task to the worker with the highest efficiency available.
  4. Calculate the total profit as the sum of the products of task difficulties and worker efficiencies.

Time complexity: O(n log n) where n is the number of tasks or workers. Space complexity: O(n) for storing the sorted tasks or workers.

:

Solution Code

import java.util.Arrays;

class Solution {
    public int maxProfitAssignment(int[] difficulty, int[] profit, int[] worker) {
        int n = difficulty.length;
        int maxProfit = 0;
        
        int[][] tasks = new int[n][2];
        for (int i = 0; i < n; i++) {
            tasks[i][0] = difficulty[i];
            tasks[i][1] = profit[i];
        }
        
        Arrays.sort(tasks, (a, b) -> a[0] - b[0]);
        Arrays.sort(worker);
        
        int i = 0, best = 0;
        for (int w : worker) {
            while (i < n && tasks[i][0] <= w) {
                best = Math.max(best, tasks[i][1]);
                i++;
            }
            maxProfit += best;
        }
        
        return maxProfit;
    }
}

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