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.
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:
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:
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:
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:
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:
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: