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:

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

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

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