Minimum Time to Complete Trips

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ArraysBinary Search

You are given an array time where time[i] denotes the time taken by the i^th bus to complete one trip.

Each bus can make multiple trips successively; that is, the next trip can start immediately after completing the current trip. Also, each bus operates independently; that is, the trips of one bus do not influence the trips of any other bus.

You are also given an integer totalTrips, which denotes the number of trips all buses should make in total. Return the minimum time required for all buses to complete at least totalTrips trips.

Example 1:

Input: time = [1,2,3], totalTrips = 5
Output: 3
Explanation:
- At time t = 1, the number of trips completed by each bus are [1,0,0]. 
  The total number of trips completed is 1 + 0 + 0 = 1.
- At time t = 2, the number of trips completed by each bus are [2,1,0]. 
  The total number of trips completed is 2 + 1 + 0 = 3.
- At time t = 3, the number of trips completed by each bus are [3,1,1]. 
  The total number of trips completed is 3 + 1 + 1 = 5.
So the minimum time needed for all buses to complete at least 5 trips is 3.

Example 2:

Input: time = [2], totalTrips = 1
Output: 2
Explanation:
There is only one bus, and it will complete its first trip at t = 2.
So the minimum time needed to complete 1 trip is 2.

Constraints:

1 <= time.length <= 10⁵
1 <= time[i], totalTrips <= 10⁷

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the number of trips required and the time each bus takes to complete a single trip? Are these integers, and can they be zero or negative?
If it's impossible to complete the required trips with the given times, what should I return?
Are there any constraints on the number of buses (the length of the `times` array)?
Can the `times` array contain duplicate values, and how would that affect the minimum time calculation?
To clarify the problem, is the time each bus takes to complete a trip constant, or can it vary? (Assuming it's constant for this problem, just confirming.)

Brute Force Solution

Approach

We want to find the smallest amount of time needed for all the vehicles to complete a certain number of trips. The brute force way is to simply try out every possible time and see if that time is good enough for all the trips to be completed.

Here's how the algorithm would work step-by-step:

Start with a very small time, like just one second.
See how many trips each vehicle can make in that one second.
Add up the number of trips all the vehicles can make together.
If the total number of trips is less than the required number of trips, then that time is not good enough. Increase the time by one second.
Repeat the process of checking the trips made for each vehicle and adding them up.
Keep increasing the time until the total number of trips made is equal to or greater than the required number of trips.
The first time you find that all vehicles together can make enough trips, that's the smallest time needed.

Code Implementation

def minimum_time_brute_force(times, total_trips):    current_time = 1
    while True:
        total_trips_completed = 0
        # Iterate through each vehicle to count its trips at current_time

        for time_for_one_trip in times:
            total_trips_completed += current_time // time_for_one_trip
        # Check if current_time is sufficient to complete all trips

        if total_trips_completed >= total_trips:
            return current_time
        # If not enough trips are completed, increment the time

        current_time += 1

Big(O) Analysis

Time Complexity

O(n*t) – The provided solution iterates through possible times, starting from 1, and incrementing it until the total number of trips made by all vehicles meets or exceeds the required number of trips. Let 'n' be the number of vehicles (length of the 'time' array) and 't' be the minimum time required to complete the trips. For each time unit, we iterate through all 'n' vehicles to calculate the number of trips each vehicle can make. In the worst case, we might have to increment the time until we reach 't'. Therefore, the outer loop iterates 't' times, and the inner loop iterates 'n' times for each value of time. Consequently, the overall time complexity is O(n*t).

Space Complexity

O(1) – The brute force solution described only uses a few integer variables to keep track of the current time and the number of trips completed. The space used by these variables does not depend on the input size (number of vehicles or required trips). Thus, the algorithm's auxiliary space complexity is constant.

Optimal Solution

Approach

The problem asks for the minimum time to complete a certain number of trips given different travel times. We can efficiently find this minimum time using a method similar to a guessing game, where we repeatedly refine our estimate until we pinpoint the exact answer.

Here's how the algorithm would work step-by-step:

Recognize that the minimum possible time is zero and the maximum possible time is when the slowest vehicle makes all the trips.
Start by guessing a time somewhere in the middle of this range.
For our guessed time, determine how many trips each vehicle can complete and add those up to get the total trips completed in that time.
Compare the total trips completed to the target number of trips. If we completed enough trips, it means our guessed time might be too high; if we didn't complete enough trips, it means our guessed time is too low.
Based on whether our guessed time was too high or too low, adjust our guess and repeat the process. If the guess was too high, guess a time in the lower half of the remaining range. If the guess was too low, guess a time in the upper half of the remaining range.
Continue adjusting our guesses, each time halving the range, until we find the smallest possible time that allows us to complete at least the target number of trips. This will be our answer.

Code Implementation

def minimum_time_to_complete_trips(times, total_trips):
    low_time = 1
    highest_time = min(times) * total_trips

    while low_time < highest_time:
        mid_time = (low_time + highest_time) // 2

        # Calculate how many trips can be completed in the mid time.
        total_possible_trips = 0
        for time in times:
            total_possible_trips += mid_time // time

        # Adjust the search range based on whether enough trips were completed
        if total_possible_trips >= total_trips:
            highest_time = mid_time
        else:
            low_time = mid_time + 1

    #At the end of the loop, low_time and highest_time converge to the answer
    return low_time

Big(O) Analysis

Time Complexity

O(n log m) – The algorithm uses binary search to find the minimum time, where m is the maximum possible time (max(times) * totalTrips). Inside the binary search loop, for each guessed time, we iterate through the 'times' array of size n to calculate the number of trips each vehicle can complete. Therefore, the time complexity is determined by the number of iterations in the binary search (log m) multiplied by the cost of each iteration which includes a loop through the array of size n. This results in a time complexity of O(n log m).

Space Complexity

O(1) – The algorithm uses a constant amount of extra space. It only needs to store a few variables like the lower bound, upper bound, middle time, and the number of trips completed for a given time. These variables consume a fixed amount of memory regardless of the input size (the number of vehicles or the target number of trips). Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty `times` array	Return 0 immediately or throw an exception, as no trips can be completed without any time values.
`trips` is zero	Return 0 immediately, as no time is needed to complete zero trips.
All `times` values are extremely large, potentially causing overflow when calculating the estimated time	Use appropriate data types (e.g., long) to prevent integer overflow during multiplication and division.
One of the `times` values is 0	Return 0 immediately if trips is zero, otherwise return maximum long value because no trips can be completed in time 0
`times` array contains negative values	Throw an IllegalArgumentException or return -1, as negative time values are invalid.
`times` array contains very small values that may lead to large intermediate calculations of number of possible trips per time unit.	Handle possible division-by-zero errors and ensure that the result of the division is accurately handled.
When the number of trips required is very large, the upper bound for the binary search could be excessively large, causing delays or overflow.	Restrict the upper bound of the search space with a reasonable estimation based on the minimum time value multiplied by the required number of trips.
The range of possible times to complete all trips could be very wide, leading to precision issues during binary search.	Ensure binary search terminates appropriately and handle rounding appropriately in the `isPossible` function to avoid off-by-one errors.

Empty `times` array

How to Handle:

Return 0 immediately or throw an exception, as no trips can be completed without any time values.

`trips` is zero

How to Handle:

Return 0 immediately, as no time is needed to complete zero trips.

All `times` values are extremely large, potentially causing overflow when calculating the estimated time

How to Handle:

Use appropriate data types (e.g., long) to prevent integer overflow during multiplication and division.

One of the `times` values is 0

How to Handle:

Return 0 immediately if trips is zero, otherwise return maximum long value because no trips can be completed in time 0

`times` array contains negative values

How to Handle:

Throw an IllegalArgumentException or return -1, as negative time values are invalid.

`times` array contains very small values that may lead to large intermediate calculations of number of possible trips per time unit.

How to Handle:

Handle possible division-by-zero errors and ensure that the result of the division is accurately handled.

When the number of trips required is very large, the upper bound for the binary search could be excessively large, causing delays or overflow.

How to Handle:

Restrict the upper bound of the search space with a reasonable estimation based on the minimum time value multiplied by the required number of trips.

The range of possible times to complete all trips could be very wide, leading to precision issues during binary search.

How to Handle:

Ensure binary search terminates appropriately and handle rounding appropriately in the `isPossible` function to avoid off-by-one errors.