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Uber Rides - Minimum Time to Complete Trips (O(n log T) Time Complexity, C++)


Uber has several cabs. The ith cab takes cabTripTime[i] minutes to complete any trip. Your task is to find the minimum time it will take Uber to get n trips completed with these cabs. You can assume there is no waiting time in-between the trips.

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

Example 1:

Input: n = 3, cabTripTime = [1, 2]
Output: 2
Explanation:
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 in 2 minutes, so the answer is 2.

Example 2:

Input: n = 10, cabTripTime = [1, 3, 5, 7]
Output: 7
Explanation:
Trips can be managed like this:
- Trip 1: 1st cab from t = 0 to t = 1;
- Trip 2: 2nd cab from t = 0 to t = 3;
- Trip 3: 3rd cab from t = 0 to t = 5.
- Trip 4: 4th cab from t = 0 to t = 7.
- Trip 5: 1st cab from t = 1 to t = 2.
- Trip 6: 1st cab from t = 2 to t = 3.
- Trip 7: 1st cab from t = 3 to t = 4.
- Trip 8: 2nd cab from t = 3 to t = 6.
- Trip 9: 1st cab from t = 4 to t = 5.
- Trip 10: 1st cab from t = 5 to t = 6.

All the trips can be completed in 7 minutes, so the answer is 7.

Note:

Your algorithm should run in O(n log T) time and use O(1) space.


Understanding the Problem

The core challenge of this problem 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 scenarios where resource allocation and time optimization are crucial, such as in ride-sharing services, task scheduling, and parallel processing.

Potential pitfalls include misunderstanding the simultaneous trip capability of cabs and not considering the optimal distribution of trips among the cabs.

Approach

To solve this problem, we can use a binary search approach on the time required to complete the trips. The idea is to determine the minimum time `T` such that the total number of trips completed by all cabs in `T` minutes is at least `n`.

Naive Solution

A naive solution would involve simulating each minute and counting the number of trips completed by all cabs. This approach is not optimal as it would take too long for large values of `n` and `T`.

Optimized Solution

The optimized solution involves using binary search to find the minimum time `T`. The steps are as follows:

  1. Initialize the search range for `T` from 1 to the maximum possible time (e.g., `n * max(cabTripTime)`).
  2. For each midpoint `mid` in the search range, calculate the total number of trips completed by all cabs in `mid` minutes.
  3. If the total number of trips is at least `n`, update the search range to find a smaller possible `T`.
  4. Otherwise, increase the search range to find a larger `T`.

Algorithm

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

  1. Initialize `left` to 1 and `right` to `n * max(cabTripTime)`.
  2. While `left` is less than or equal to `right`:
    • Calculate `mid` as the average of `left` and `right`.
    • Calculate the total number of trips completed by all cabs in `mid` minutes.
    • If the total number of trips is at least `n`, update `right` to `mid - 1`.
    • Otherwise, update `left` to `mid + 1`.
  3. The minimum time `T` is `left`.

Code Implementation


#include <vector>
#include <algorithm>
#include <climits>

using namespace std;

// Function to calculate the total number of trips completed in 'time' minutes
long long totalTrips(const vector<int>& cabTripTime, long long time) {
    long long trips = 0;
    for (int t : cabTripTime) {
        trips += time / t;
    }
    return trips;
}

int minimumTimeToCompleteTrips(int n, vector<int>& cabTripTime) {
    // Initialize the search range
    long long left = 1;
    long long right = (long long)n * *max_element(cabTripTime.begin(), cabTripTime.end());
    long long result = right;

    // Binary search to find the minimum time
    while (left <= right) {
        long long mid = left + (right - left) / 2;
        if (totalTrips(cabTripTime, mid) >= n) {
            result = mid;
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }

    return result;
}

// Example usage
int main() {
    vector<int> cabTripTime1 = {1, 2};
    int n1 = 3;
    cout << "Minimum time to complete " << n1 << " trips: " << minimumTimeToCompleteTrips(n1, cabTripTime1) << endl;

    vector<int> cabTripTime2 = {1, 3, 5, 7};
    int n2 = 10;
    cout << "Minimum time to complete " << n2 << " trips: " << minimumTimeToCompleteTrips(n2, cabTripTime2) << endl;

    return 0;
}

Complexity Analysis

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.

Edge Cases

Potential edge cases include:

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

Testing

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

Testing frameworks such as Google Test can be used to automate and validate the test cases.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Conclusion

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 C++. Understanding and solving such problems is crucial for optimizing resource allocation and time management in various applications.

Additional Resources

For further reading and practice, consider the following resources: