LeetCode-in-Java

1870. Minimum Speed to Arrive on Time

Medium

You are given a floating-point number hour, representing the amount of time you have to reach the office. To commute to the office, you must take n trains in sequential order. You are also given an integer array dist of length n, where dist[i] describes the distance (in kilometers) of the ith train ride.

Each train can only depart at an integer hour, so you may need to wait in between each train ride.

• For example, if the 1st train ride takes 1.5 hours, you must wait for an additional 0.5 hours before you can depart on the 2nd train ride at the 2 hour mark.

Return the minimum positive integer speed (in kilometers per hour) that all the trains must travel at for you to reach the office on time, or -1 if it is impossible to be on time.

Tests are generated such that the answer will not exceed 107 and hour will have at most two digits after the decimal point.

Example 1:

Input: dist = [1,3,2], hour = 6

Output: 1

Explanation: At speed 1:

• The first train ride takes 1/1 = 1 hour.

• Since we are already at an integer hour, we depart immediately at the 1 hour mark. The second train takes 3/1 = 3 hours.

• Since we are already at an integer hour, we depart immediately at the 4 hour mark. The third train takes 2/1 = 2 hours.

• You will arrive at exactly the 6 hour mark.

Example 2:

Input: dist = [1,3,2], hour = 2.7

Output: 3

Explanation: At speed 3:

• The first train ride takes 1/3 = 0.33333 hours.

• Since we are not at an integer hour, we wait until the 1 hour mark to depart. The second train ride takes 3/3 = 1 hour.

• Since we are already at an integer hour, we depart immediately at the 2 hour mark. The third train takes 2/3 = 0.66667 hours.

• You will arrive at the 2.66667 hour mark.

Example 3:

Input: dist = [1,3,2], hour = 1.9

Output: -1

Explanation: It is impossible because the earliest the third train can depart is at the 2 hour mark.

Constraints:

• n == dist.length
• 1 <= n <= 105
• 1 <= dist[i] <= 105
• 1 <= hour <= 109
• There will be at most two digits after the decimal point in hour.

Solution

@SuppressWarnings("java:S2184")
public class Solution {
    public int minSpeedOnTime(int[] dist, double hour) {
        int n = dist.length;
        return fmin(dist, n, hour);
    }

    private boolean check(int[] dist, int n, double h, int spe) {
        double cost = 0;
        for (int i = 0; i < n - 1; i++) {
            // same as ceil(doubleTime/doubleSpeed)
            cost += (dist[i] - 1) / spe + 1;
        }
        cost += (double) dist[n - 1] / (double) spe;
        return cost <= h;
    }

    private int fmin(int[] dist, int n, double h) {
        if (h + 1 <= n) {
            return -1;
        }
        int max = fmax(dist) * 100;
        int lo = 1;
        int hi = max;
        while (lo < hi) {
            int mid = (lo + hi) / 2;
            // speed of mid is possible, move to left side
            if (check(dist, n, h, mid)) {
                hi = mid;
            } else {
                // need higher speed, move to right side
                lo = mid + 1;
            }
        }
        return lo;
    }

    private int fmax(int[] arr) {
        int res = arr[0];
        for (int num : arr) {
            res = Math.max(res, num);
        }
        return res;
    }
}

This site is open source. Improve this page.