安排工作以达到最大收益

/*
 * Problem statement:
 * You have n jobs and m workers. Given three arrays: difficulty, profit, and worker, where:
 * difficulty[i] indicates the difficulty of the ith job, profit[i] indicates the profit of the ith job.
 * worker[i] is the ability of the ith worker, meaning that this worker can only complete a job with difficulty less than or equal to worker[i].
 * Each worker can be assigned at most one job, but one job can be completed multiple times.
 * Return the maximum profit we can achieve after assigning workers to jobs.
 */
import java.util.Arrays;

public class MaxProfitAssignmentStream {
    public int maxProfitAssignment(int[] difficulty, int[] profit, int[] worker) {
        // Combine difficulty and profit arrays into a single 2D array
        int[][] jobs = new int[difficulty.length][2];
        for (int i = 0; i < difficulty.length; i++) {
            jobs[i][0] = difficulty[i];
            jobs[i][1] = profit[i];
        }
        // Sort jobs by difficulty and worker by ability
        Arrays.sort(jobs, (a, b) -> a[0] - b[0]);
        Arrays.sort(worker);
        // Use streams to find the maximum profit
        int[] maxProfit = {0};
        int[] best = {0};
        int[] jobIndex = {0};
        Arrays.stream(worker).forEach(w -> {
            while (jobIndex[0] < jobs.length && w >= jobs[jobIndex[0]][0]) {
                best[0] = Math.max(best[0], jobs[jobIndex[0]][1]);
                jobIndex[0]++;
            }
            maxProfit[0] += best[0];
        });
        return maxProfit[0];
    }
}