Formatted question description: https://leetcode.ca/all/1870.html

# 1870. Minimum Speed to Arrive on Time

Medium

## Description

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 i-th 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 10^7 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 <= 10^5
• 1 <= dist[i] <= 10^5
• 1 <= hour <= 10^9
• There will be at most two digits after the decimal point in hour.

## Solution

The maximum possible speed is 10^7. If it is still impossible to arrive on time using the maximum possible speed, return -1. Then let high = 10000000 and low = 1, and use binary search to find the minimum speed to arrive on time.

class Solution {
public int minSpeedOnTime(int[] dist, double hour) {
int high = 10000000;
if (!canArrive(dist, hour, high))
return -1;
int low = 1;
while (low < high) {
int mid = (high - low) / 2 + low;
if (canArrive(dist, hour, mid))
high = mid;
else
low = mid + 1;
}
return low;
}

public boolean canArrive(int[] dist, double hour, int speed) {
double total = 0;
int length = dist.length;
for (int i = 0; i < length; i++) {
if (i == length - 1)
total += 1.0 * dist[i] / speed;
else {
int curr = dist[i] / speed;
if (dist[i] % speed != 0)
curr++;
total += curr;
}
}