Problem Description

In this problem, we are given an array time that represents the time each bus takes to complete a single trip. Importantly, the buses can each make as many trips as needed back-to-back, without any downtime in between successive trips. Moreover, the buses operate independently from one another, meaning that the trips one bus makes have no effect on another's trips.

We also receive an integer totalTrips, which signifies the collective number of trips that all buses combined must complete. The task is to determine the minimum amount of time required for all buses to collectively complete at least totalTrips trips. Keep in mind that it's the combined trips of all buses that we are interested in, not just the trips made by one bus.

Intuition

The solution revolves around the concept of binary search. Initially, we can think that the minimum time required for all buses to complete the required number of trips would be if all buses operated at their fastest possible times, which is why we use the minimum bus time multiplied by the totalTrips as an upper bound for the time needed.

Given that the buses operate independently, we need to find the point at which the sum of trips made by all buses hits at least totalTrips. To do this efficiently, we use binary search instead of checking each possible time sequentially.

Binary search works here because as time increases, the number of trips that can be completed is non-decreasing - this is a monotonic relationship, which is a requirement for binary search to be applicable.

The key insight behind the solution is that for a given amount of time t, we can calculate the total number of trips completed by all buses by summing up t // time[i] for each bus i. If this total number of trips is less than totalTrips, we know we need more time, and if it's more, we have a potential solution but we'll continue to search for a smaller time value that still meets the totalTrips requirement.

The bisect_left function from Python's standard library is then used to find the smallest t where the calculated number of trips is at least totalTrips. It does this by looking for the leftmost insertion point to maintain sorted order.

Learn more about Binary Search patterns.

Solution Approach

The solution leverages binary search, an efficient algorithm for finding an item from a sorted collection by repeatedly halving the search interval. In this problem, although the starting time array is not inherently sorted, we are searching through a range of time from 0 to mx, where mx is the product of the minimum value in the time array and totalTrips. This range of time is conceptually sorted because, as mentioned earlier, the number of total trips completed by all buses increases monotonically as time increases.

The Python bisect_left function is used in this context to perform the binary search. The bisect_left function requires a range or a sorted list to search within, a target value to find, and an optional key function to transform the elements before comparison.

In this problem, our range represents all possible times from 0 to mx. The target we are seeking is totalTrips, our criterion for the minimum trips required.

The key function is particularly significant here. For every mid-value time x during the binary search, the key function calculates the total number of trips that all buses would have completed by that time using the expression sum(x // v for v in time) where v is each individual time from the time list. The // operator performs integer division, yielding the number of trips a single bus completes in time x.

The process goes as follows:

• If the calculated total trips for the mid time value is less than totalTrips, it means more time is needed for the buses to complete the required number of trips, so the search continues to the right half of the current range.
• If the total is greater or equal, the mid value is a potential solution, but we check to the left half of the range to find if there is a smaller viable time.

This continues until the binary search narrows down to the minimum time where the buses can collectively complete totalTrips. That is the value returned by the bisect_left function as a result, hence providing us with the minimum time needed to achieve at least totalTrips trips by all buses.

Here is the important part of the solution code for reference, focusing on the applied binary search mechanism:

class Solution:
    def minimumTime(self, time: List[int], totalTrips: int) -> int:
        mx = min(time) * totalTrips
        return bisect_left(
            range(mx), totalTrips, key=lambda x: sum(x // v for v in time)
        )

No additional data structures are needed beyond the input time list and the range object created for the search interval. The binary search pattern itself acts as the algorithmic data structure guiding the search for the correct minimum time.

Example Walkthrough

Let's walk through a simple example to illustrate the solution approach.

Suppose time = [3, 6, 8] represents the time taken by three buses to complete a single trip. We need to find the minimum amount of time required for these buses to complete at least totalTrips = 7 trips in total.

First, we determine an upper bound for the binary search based on the fastest bus. The minimum time in time is 3, so the maximum conceivable time would be 3 * 7 = 21. This is because if only the fastest bus were running, it would take 21 units of time to complete 7 trips.

Next, we use binary search to find the minimum time. The range for binary search is from 0 to 21:

1. In the first iteration, the midpoint (mid) of 0 and 21 is 10. We check how many total trips can be completed by all the buses in 10 units of time.

   • Bus 1 can make 10 // 3 = 3 trips.
   • Bus 2 can make 10 // 6 = 1 trip.
   • Bus 3 can make 10 // 8 = 1 trip.
   • So, the total is 3 + 1 + 1 = 5 trips, which is less than 7. We need more time.

2. The next mid will be the midpoint between 11 and 21, which is 16.

   • Bus 1 can make 16 // 3 = 5 trips.
   • Bus 2 can make 16 // 6 = 2 trips.
   • Bus 3 can make 16 // 8 = 2 trips.
   • Now, the total is 5 + 2 + 2 = 9 trips, which is more than 7. We have a potential solution but we'll continue to search in case there's a lower value that also works.

3. The next midpoint will be between 11 and 15, giving us 13.

   • Bus 1 can make 13 // 3 = 4 trips.
   • Bus 2 can make 13 // 6 = 2 trips.
   • Bus 3 can make 13 // 8 = 1 trip.
   • The total is 4 + 2 + 1 = 7 trips, which is exactly what we need.

4. Even though we've found that 13 minutes would work, we need to make sure there isn't a smaller value that also meets the totalTrips requirement, so we look to the left, between 11 and 12.

5. As we proceed with the search, we find that 12 minutes won't suffice for 7 trips (it only yields 6 trips in total), so we finally conclude that the minimum time needed is 13.

Following the binary search pattern, we get the minimum time required for all buses to collectively complete at least totalTrips trips which, in this example, is 13 units of time.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code uses a binary search approach to find the minimum time required to make totalTrips using a list of time where each time[i] represents the time needed to complete one trip.

The time complexity of the binary search is O(log(max_time)), where max_time = min(time) * totalTrips. This represents the maximum amount of time it might take if the slowest worker alone had to complete all totalTrips.

Within each step of the binary search, the code executes a sum operation that runs in O(n) time, where n is the number of workers (or the length of the time list). This sum operation calculates the total number of trips completed within a given time frame.

Therefore, the total time complexity of the code is O(n * log(max_time)).

Space Complexity

The space complexity of the code is O(1) since it only uses constant extra space for variables such as mx and the lambda function. There is no use of any additional data structures that grow with the input size.

Learn more about how to find time and space complexity quickly using problem constraints.