Uber has several cabs. The **i ^{th}** cab takes

**Note:** Different cabs can take trips simultaneously (at any time, there can be more than one cab with an ongoing trip)

**n**: An integer representing the number of trips.**cabTripTime**: An array of integers where**cabTripTime[i]**represents the time taken by the**i**cab to complete a trip.^{th}

- An integer representing the minimum time required to complete
**n**trips.

- 1 ≤ n ≤ 10
^{9} - 1 ≤ cabTripTime.length ≤ 10
^{5} - 1 ≤ cabTripTime[i] ≤ 10
^{7}

**Example 1:**

Input:n= 3,cabTripTime=`[1, 2]`

Output:2Explanation:Trips can be managed like this: - Trip 1: first cab from t = 0 to t = 1; - Trip 2: second cab from t = 0 to t = 2; - Trip 3: first cab from t = 1 to t = 2. All the trips can be completed in2minutes, so the answer is2.

**Example 2:**

Input:n= 10,cabTripTime=`[1, 3, 5, 7]`

Output:7Explanation:Trips can be managed like this: - Trip 1: 1^{st}cab from t = 0 to t = 1; - Trip 2: 2^{nd}cab from t = 0 to t = 3; - Trip 3: 3^{rd}cab from t = 0 to t = 5. - Trip 4: 4^{th}cab from t = 0 to t = 7. - Trip 5: 1^{st}cab from t = 1 to t = 2. - Trip 6: 1^{st}cab from t = 2 to t = 3. - Trip 7: 1^{st}cab from t = 3 to t = 4. - Trip 8: 2^{nd}cab from t = 3 to t = 6. - Trip 9: 1^{st}cab from t = 4 to t = 5. - Trip 10: 1^{st}cab from t = 5 to t = 6. All the trips can be completed in7minutes, so the answer is7.

The core challenge is to determine the minimum time required to complete a given number of trips using multiple cabs, each with a different trip completion time. This problem is significant in optimizing resource allocation and scheduling in real-time systems like ride-sharing services.

Potential pitfalls include misunderstanding that cabs can operate simultaneously and not accounting for the varying trip times of different cabs.

To solve this problem, we need to determine the minimum time required to complete **n** trips using the given cabs. A naive approach would be to simulate each trip, but this would be inefficient for large inputs. Instead, we can use a binary search approach to find the minimum time efficiently.

The naive solution involves simulating each trip and assigning it to the next available cab. This approach is not optimal as it has a time complexity of O(n * m), where m is the number of cabs.

We can use a binary search to find the minimum time. The idea is to search for the minimum time **T** such that the total number of trips completed by all cabs in time **T** is at least **n**.

1. Initialize the search range: **left** = 1 and **right** = max(cabTripTime) * n.

2. Perform binary search within this range:

- Calculate the mid-point
**mid**= (left + right) / 2. - Calculate the total number of trips that can be completed in
**mid**minutes. - If the total number of trips is at least
**n**, update**right**to**mid**. - Otherwise, update
**left**to**mid + 1**.

3. The minimum time required will be the value of **left** after the binary search completes.

Here is a step-by-step breakdown of the binary search algorithm:

- Initialize
**left**to 1 and**right**to max(cabTripTime) * n. - While
**left**is less than**right**: - Calculate
**mid**as the average of**left**and**right**. - Calculate the total number of trips that can be completed in
**mid**minutes. - If the total number of trips is at least
**n**, update**right**to**mid**. - Otherwise, update
**left**to**mid + 1**. - Return
**left**as the minimum time required.

```
public class UberRides {
public static int minTimeToCompleteTrips(int n, int[] cabTripTime) {
// Initialize the search range
int left = 1;
int right = Integer.MAX_VALUE;
// Binary search to find the minimum time
while (left < right) {
int mid = left + (right - left) / 2;
if (canCompleteTrips(n, cabTripTime, mid)) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
private static boolean canCompleteTrips(int n, int[] cabTripTime, int time) {
long totalTrips = 0;
for (int tripTime : cabTripTime) {
totalTrips += time / tripTime;
if (totalTrips >= n) {
return true;
}
}
return totalTrips >= n;
}
public static void main(String[] args) {
int n1 = 3;
int[] cabTripTime1 = {1, 2};
System.out.println(minTimeToCompleteTrips(n1, cabTripTime1)); // Output: 2
int n2 = 10;
int[] cabTripTime2 = {1, 3, 5, 7};
System.out.println(minTimeToCompleteTrips(n2, cabTripTime2)); // Output: 7
}
}
```

The time complexity of the binary search approach is O(n log T), where T is the maximum possible time. The space complexity is O(1) as we are using a constant amount of extra space.

Potential edge cases include:

- All cabs have the same trip time.
- Only one cab is available.
- The number of trips is very large.

Each algorithm handles these edge cases by ensuring the binary search covers the entire range of possible times and correctly calculates the total number of trips for any given time.

To test the solution comprehensively, consider the following test cases:

- Simple cases with small values of
**n**and**cabTripTime**. - Cases with large values of
**n**to test performance. - Edge cases with only one cab or all cabs having the same trip time.

Use a variety of test cases to ensure the solution is robust and handles all scenarios correctly.

When approaching such problems, consider the following tips:

- Break down the problem into smaller parts and understand the requirements.
- Consider different approaches and their time complexities.
- Use binary search for optimization problems involving a range of values.
- Practice solving similar problems to improve problem-solving skills.

In this blog post, we discussed how to solve the problem of finding the minimum time to complete a given number of trips using multiple cabs with different trip times. We explored a binary search approach to optimize the solution and provided a detailed explanation of the algorithm and its implementation in Java. Understanding and solving such problems is crucial for optimizing resource allocation and scheduling in real-time systems.

For further reading and practice problems related to this topic, consider the following resources:

- LeetCode - A platform for practicing coding problems.
- GeeksforGeeks - A comprehensive resource for learning algorithms and data structures.
- Coursera - Online courses on algorithms and problem-solving.