Minimum Time to Complete Trips

class Solution {
public:
    bool sufficientTripTime(vector<int>& durations, long long maxDuration, int requiredTrips) {
        long long tripsCount = 0;
        for (int duration : durations) {
            tripsCount += maxDuration / duration;
        }
        return tripsCount >= requiredTrips;
    }
    long long minTimeToCompleteTrips(vector<int>& durations, int requiredTrips) {
        long long minTime = 1, maxTime = 1LL * *max_element(durations.begin(), durations.end()) * requiredTrips;

        while (minTime < maxTime) {
            long long middleTime = (minTime + maxTime) / 2;
            if (sufficientTripTime(durations, middleTime, requiredTrips)) {
                maxTime = middleTime;
            } else {
                minTime = middleTime + 1;
            }
        }
        return minTime;
    }
};

You want to find the minimum time required to complete a given number of trips using vehicles with specified durations for single trips. Below is the implementation in C++ that will achieve this.

• Start by writing the helper function sufficientTripTime, which checks if a given maximum duration allows all trips to be made. It calculates the total number of trips that can be completed within this time by iterating over each vehicle's duration, adding up how many trips each can make within the maxDuration.

• Implement the main function minTimeToCompleteTrips to determine the minimum required time to complete the specified number of trips using binary search:

  1. Initialize minTime to 1 and maxTime to the product of the maximum single trip duration and the number of required trips.
  2. Execute a binary search on time:
     • Calculate the midpoint middleTime as the average of minTime and maxTime.
     • Use the sufficientTripTime function to check if the current middleTime is sufficient to complete the required trips.
     • If true, update maxTime to middleTime to search for a possibly smaller valid time.
     • If false, adjust minTime to middleTime + 1 to omit non-feasible durations.
  3. The search loop continues until minTime equals maxTime, the point at which minTime will represent the minimum time required to complete the trips.

This binary search approach efficiently narrows down the minimal time needed by reducing the search range, leveraging the fact that if a certain time is sufficient to complete the trips, any time longer than that will also suffice (and vice versa).

class Solution {
    public boolean canCompleteTrips(int[] durations, long maxDuration, int requiredTrips) {
        long trips = 0;
        for (int duration : durations) 
            trips += maxDuration / duration;
        return trips >= requiredTrips;
    }
    
    public long findLeastTime(int[] durations, int requiredTrips) {
        int longest = 0;
        for (int t : durations) {
            longest = Math.max(longest, t);
        }
        long low = 1, high = (long) longest * requiredTrips;
        
        while (low < high) {
            long mid = (low + high) / 2;
            if (canCompleteTrips(durations, mid, requiredTrips)) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
        return low;
    }
}

The solution presented in Java for the problem of finding the minimum time to complete a specified number of trips using different durations focuses on an efficient approach using binary search.

• The solution comprises two primary functions:
  • canCompleteTrips(int[] durations, long maxDuration, int requiredTrips) verifies if the total required trips can be accomplished within the provided maximum duration.
  • findLeastTime(int[] durations, int requiredTrips) calculates the minimum time to complete the required trips.

1. The canCompleteTrips method computes the total number of trips possible by iterating through each trip duration and summing how many trips can be made within the maxDuration using integer division.
2. In the findLeastTime method:
   • Identify the maximum trip duration in the array, to establish a reasonable upper limit for the binary search.
   • Set initial boundaries for binary search; low starts at 1 and high is calculated as the product of the longest duration and required trips.
   • Through binary search, determine the smallest maximum time interval (low) where the number of trips meets or exceeds the required count.

This binary search strategy efficiently narrows down the potential minimal duration necessary to meet the required number of trips, ensuring an optimal and quick solution.

class Solution:
    def findMinTime(self, durations: List[int], requiredTrips: int) -> int:
        # Set initial search boundaries
        low, high = 1, max(durations) * requiredTrips

        def sufficientTime(mid_time):
            current_trips = 0
            for duration in durations:
                current_trips += mid_time // duration
            return current_trips >= requiredTrips

        # Perform binary search to find the lowest sufficient time
        while low < high:
            midpoint = (low + high) // 2
            if sufficientTime(midpoint):
                high = midpoint
            else:
                low = midpoint + 1
        return low

The following Python3 solution addresses the problem of finding the minimum time required to complete a specified number of trips using different vehicles that operate at varying speeds. This problem is efficiently solved using a combination of binary search and a helper function.

The findMinTime method accepts two parameters: an array of integers durations where each integer represents the time taken by a vehicle to complete one trip, and an integer requiredTrips which is the total number of trips that must be completed across all vehicles.

Here’s a stepwise explanation of the code:

1. Define the search range's initial bounds:

   • low is initialized as 1 — the minimum conceivable time is often considered as one time unit since no trip can be completed in zero time.
   • high is initialized to the product of the maximum single trip duration in durations and requiredTrips — this assumes a worst-case scenario where the slowest vehicle might conceivably take all the trips sequentially, thus providing an upper bound on the search space.

2. Implement the sufficientTime(mid_time) helper function to determine whether all required trips can be completed within mid_time:

   • Initialize current_trips to zero.
   • Iterate through the durations array, summing up the number of trips that each vehicle can complete within mid_time.
   • Return a boolean indicating if the number of computed trips is adequate (>= requiredTrips).

3. Conduct a binary search using these boundaries to home in on the minimal time:

   • While the lower bound (low) is less than the upper (high), calculate the midpoint and check if it suffices via sufficientTime(midpoint).
   • If it is sufficient, adjust the upper bound (high) to the midpoint, effectively narrowing down the potential minimum time. Otherwise, adjust the lower bound to just above the midpoint (low = midpoint + 1).

4. Once the loop completes, low will be at the minimal sufficient time value, and it is returned.

This approach is efficient, leveraging a binary search that typically completes in logarithmic time relative to the size of the input, avoiding the need for brute force examination of every possible time duration from 1 to high. Armed with this solution, you have the capacity to optimize travel or operational schedules where time efficiency is paramount.