完成所有任务的最少时间

/*
 * Problem Description: 
 * You have a computer that can run an infinite number of tasks simultaneously. Given a 2D integer array 'tasks' where tasks[i] = [start_i, end_i, duration_i] indicates that the i-th task needs to run for 'duration_i' integer time points within the closed interval [start_i, end_i]. You can turn the computer on when a task needs to run and turn it off when idle. Return the minimum number of seconds the computer needs to run to complete all tasks.
 * 
 * Approach: 
 * 1. Sort the tasks by their end time to ensure we can use the latest possible time points. 
 * 2. Use a priority queue to track the current running tasks and their end times. 
 * 3. Traverse through the tasks, updating the priority queue and calculating the required run times.
 */

import java.util.*;
import java.util.stream.*;

public class MinRunTimeStream {
    public int minRunTime(int[][] tasks) {
        Arrays.sort(tasks, Comparator.comparingInt(task -> task[1]));
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        int[] currentTime = {0};
        Arrays.stream(tasks).forEach(task -> {
            int start = task[0];
            int end = task[1];
            int duration = task[2];
            IntStream.range(0, duration).forEach(i -> {
                if (!pq.isEmpty() && pq.peek() < start) {
                    currentTime[0] = pq.poll();
                } else {
                    currentTime[0] = Math.max(currentTime[0], start);
                }
                pq.offer(currentTime[0]++);
            });
        });
        return pq.size();
    }

    public static void main(String[] args) {
        MinRunTimeStream mrt = new MinRunTimeStream();
        int[][] tasks = {{2, 3, 1}, {4, 5, 1}, {1, 5, 2}};
        System.out.println(mrt.minRunTime(tasks)); // Output: 2
    }
}