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.
Your algorithm should run in O(n log T) time and use O(1) space.
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.
To solve this problem, we need to think about how to distribute the trips among the cabs in such a way that the total time is minimized. A naive approach would be to simulate each trip and assign it to the next available cab, but this would be inefficient for large inputs.
Instead, we can use a binary search approach to find the minimum time required. The idea is to search for the minimum possible time (let's call it T) such that the sum of the trips completed by all cabs in time T is at least n.
1. Initialize the search range: The minimum possible time is 1 minute, and the maximum possible time is n * max(cabTripTime).
2. Perform binary search within this range to find the minimum time T that allows completing at least n trips.
3. For each midpoint mid in the binary search, calculate the total number of trips that can be completed by all cabs in mid minutes.
4. Adjust the search range based on whether the total number of trips is greater than or equal to n.
Here is a step-by-step breakdown of the binary search algorithm:
// Function to calculate the total number of trips that can be completed in 'time' minutes
function totalTrips(cabTripTime, time) {
let trips = 0;
for (let i = 0; i < cabTripTime.length; i++) {
trips += Math.floor(time / cabTripTime[i]);
}
return trips;
}
// Function to find the minimum time to complete 'n' trips
function minimumTime(n, cabTripTime) {
let left = 1;
let right = n * Math.max(...cabTripTime);
let result = right;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (totalTrips(cabTripTime, mid) >= n) {
result = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return result;
}
// Example usage:
const n1 = 3;
const cabTripTime1 = [1, 2];
console.log(minimumTime(n1, cabTripTime1)); // Output: 2
const n2 = 10;
const cabTripTime2 = [1, 3, 5, 7];
console.log(minimumTime(n2, cabTripTime2)); // Output: 7
The time complexity of this approach is O(n log T), where T is the maximum possible time. The space complexity is O(1) since we are using a constant amount of extra space.
Potential edge cases include:
n is very large.Each of these cases should be tested to ensure the algorithm handles them correctly.
To test the solution comprehensively, consider the following test cases:
n and cabTripTime.n and varying cabTripTime.Using a testing framework like Jest can help automate and validate these test cases.
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, code implementation, and complexity analysis. Understanding and solving such problems is crucial for optimizing resource allocation and improving efficiency in various applications.
For further reading and practice, consider the following resources: