leetcode - 1870. Minimum Speed to Arrive on Time

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 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:

复制代码
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:

复制代码
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:

复制代码
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:

复制代码
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

Since speed is positive integer, and for each speed we know whether this speed can get us to the destination within give time or not, we could use binary search to find the optimal speed. Try to find the smallest speed that is valid.

To start searching, firstly we need to determine the search range. I used some maths to calculate the range, but actually use [1, 10^7] would be easier.

Time complexity: o ( log ⁡ n ) o(\log n) o(logn), where n ∈ [ 1 , 1 0 7 ] n \in [1, 10^7] n∈[1,107]

Space complexity: o ( 1 ) o(1) o(1)

Code

python3 复制代码
class Solution:
    def is_valid(self, speed: int, dist: list, hour: float) -> bool:
        """
        Decide whether using this speed is valid or not
        """
        sum_of_time = 0
        for each_dist in dist[:-1]:
            sum_of_time += math.ceil(each_dist / speed)
        sum_of_time += dist[-1] / speed
        return True if sum_of_time <= hour else False


    def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
        if hour - len(dist) + 1 <= 0:
            return -1

        import math
        left = max(math.floor(sum(dist) / hour), 1)
        right = max(max(dist), math.ceil(dist[-1] / (hour - len(dist) + 1)))
        while left < right:
            mid = (left + right) >> 1
            if not self.is_valid(mid, dist, hour):
                left = mid + 1
            else:
                right = mid
        mid = (left + right) >> 1
        if self.is_valid(mid, dist, hour):
            return mid
        return -1
相关推荐
你撅嘴真丑13 小时前
第九章-数字三角形
算法
uesowys14 小时前
Apache Spark算法开发指导-One-vs-Rest classifier
人工智能·算法·spark
ValhallaCoder14 小时前
hot100-二叉树I
数据结构·python·算法·二叉树
董董灿是个攻城狮14 小时前
AI 视觉连载1:像素
算法
智驱力人工智能14 小时前
小区高空抛物AI实时预警方案 筑牢社区头顶安全的实践 高空抛物检测 高空抛物监控安装教程 高空抛物误报率优化方案 高空抛物监控案例分享
人工智能·深度学习·opencv·算法·安全·yolo·边缘计算
孞㐑¥15 小时前
算法——BFS
开发语言·c++·经验分享·笔记·算法
月挽清风15 小时前
代码随想录第十五天
数据结构·算法·leetcode
XX風15 小时前
8.1 PFH&&FPFH
图像处理·算法
NEXT0616 小时前
前端算法:从 O(n²) 到 O(n),列表转树的极致优化
前端·数据结构·算法
代码游侠16 小时前
学习笔记——设备树基础
linux·运维·开发语言·单片机·算法