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
Yiyaoshujuku12 分钟前
疾病的发病率、发病人数、患病率、患病人数、死亡率、死亡人数查询网站及数据库
数据库·人工智能·算法
wen__xvn13 分钟前
基础算法集训第18天:深度优先搜索
算法·深度优先·图论
jiang_changsheng25 分钟前
comfyui节点插件笔记总结新增加
人工智能·算法·计算机视觉·comfyui
TracyCoder12325 分钟前
LeetCode Hot100(7/100)—— 3. 无重复字符的最长子串
算法·leetcode
重生之我是Java开发战士29 分钟前
【优选算法】双指针法:移动0,复写0,快乐数,盛水最多的容器,有效三角形个数,二三四数之和
算法
Jeremy爱编码36 分钟前
计算机操作系统
职场和发展
客卿1231 小时前
力扣二叉树简单题整理--(包含常用语法的讲解)
算法·leetcode·职场和发展
hrrrrb1 小时前
【算法设计与分析】递归与分治策略
算法
鱼跃鹰飞1 小时前
大厂面试真题-说说kafka消费端幂等性?
面试·职场和发展·kafka
We་ct1 小时前
LeetCode 28. 找出字符串中第一个匹配项的下标:两种实现与深度解析
前端·算法·leetcode·typescript