[1870. Minimum Speed to Arrive on Time](https://leetcode.com/problems/minimum-speed-to-arrive-on-time/)

### 題目描述

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 1^st^ train ride takes `1.5` hours, you must wait for an additional `0.5` hours before you can depart on the 2^nd^ 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**.

### 範例

**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`.

### 解答
#### C++

class Solution {
public:
    int minSpeedOnTime(vector<int>& dist, double timeLimit) {
        // ios_base::sync_with_stdio(0); cin.tie(0);

        // const int MAX_SPEED = *max_element(dist.begin(), dist.end()) + 1;
        const int MAX_SPEED = 1e7 + 1;
        int left = 1, right = MAX_SPEED;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (speedOnTime(dist, timeLimit, mid)) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return right == MAX_SPEED ? -1 : right;
    }

    bool speedOnTime(vector<int>& dist, double timeLimit, int speed) {
        double timeSpend = 0;
        for (int i = 0; i < dist.size() - 1; i ++) {
            timeSpend += ceil((double)dist[i] / speed);
        }
        timeSpend += (double) dist.back() / speed;
        // printf("speed=%d\ttimeSpend=%f\ttimeLimit=%f\n", speed, timeSpend, timeLimit);
        return timeSpend <= timeLimit;
    }
};

需要注意以下測資
dist=`[1,1,100000]`
hour=`2.01`

> [name=Jerry Wu][time=26 July, 2023]

#### Javascript

function minSpeedOnTime(dist, hour) {
  let max = 1e7;
  let min = 1;
  const n = dist.length;
  let result = -1;

  while (max >= min) {
    const mid = (max + min) >> 1;
    let time = 0;
    for (let i = 0; i < n - 1; i++) {
      time += Math.ceil(dist[i] / mid);
    }
    time += dist[n - 1] / mid;

    if (time > hour) {
      min = mid + 1;
    } else {
      result = mid;
      max = mid - 1;
    }
  }

  return result;
}

> 原本的 max 是用dist中的最大值，結果錯了樓上說要注意的測資ㄏㄏ
> [name=Marsgoat][time=26 July, 2023]

### Reference

[回到題目列表](https://hackmd.io/@Marsgoat/leetcode_every_day)