车队 II

/*
 * Approach:
 * 1. Similar to the Java solution but using Java Stream API for a more functional approach.
 * 2. We map each pair of consecutive cars to their meeting time or -1 if they don't meet.
 */
import java.util.stream.IntStream;
import java.util.stream.DoubleStream;

public class CarFleetTimeStream {
    public static double[] getFleetMeetingTimes(int[][] cars) {
        int n = cars.length;
        return IntStream.range(0, n)
                .mapToDouble(i -> {
                    if (i == n - 1) return -1.0;
                    int pos1 = cars[i][0], speed1 = cars[i][1];
                    int pos2 = cars[i + 1][0], speed2 = cars[i + 1][1];
                    return speed1 <= speed2 ? -1.0 : (double) (pos2 - pos1) / (speed1 - speed2);
                })
                .toArray();
    }
}