准时抵达会议现场的最小跳过休息次数

/*
 * Problem: Similar to the Java solution, but using Java Streams where applicable.
 * Idea: Utilize streams for operations like array initialization and minimum calculation.
 */

import java.util.Arrays;
import java.util.stream.IntStream;

public class MeetingSchedulerStream {
    public int minSkips(int[] dist, int speed, int hoursBefore) {
        int n = dist.length;
        double[][] time = new double[n + 1][n + 1];
        Arrays.stream(time).forEach(row -> Arrays.fill(row, Double.MAX_VALUE));
        time[0][0] = 0;

        for (int i = 1; i <= n; i++) {
            double t = (double) dist[i - 1] / speed;
            for (int j = 0; j <= i; j++) {
                if (j < i) {
                    time[i][j] = Math.min(time[i][j], Math.ceil(time[i - 1][j]) + t);
                }
                if (j > 0) {
                    time[i][j] = Math.min(time[i][j], time[i - 1][j - 1] + t);
                }
            }
        }

        return IntStream.range(0, n + 1)
                         .filter(j -> time[n][j] <= hoursBefore)
                         .findFirst()
                         .orElse(-1);
    }
}