1870. 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 10⁷ 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 <= 10⁵
• 1 <= dist[i] <= 10⁵
• 1 <= hour <= 10⁹
• 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

Jerry Wu26 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
Marsgoat26 July, 2023

Reference

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