You are given an integer array ranks representing the ranks of some mechanics. ranksi is the rank of the ith mechanic. A mechanic with a rank r can repair n cars in r * n2 minutes.
You are also given an integer cars representing the total number of cars waiting in the garage to be repaired.
Return the minimum time taken to repair all the cars.
Note: All the mechanics can repair the cars simultaneously.
Example 1:
Input: ranks = [4,2,3,1], cars = 10 Output: 16 Explanation: - The first mechanic will repair two cars. The time required is 4 * 2 * 2 = 16 minutes. - The second mechanic will repair two cars. The time required is 2 * 2 * 2 = 8 minutes. - The third mechanic will repair two cars. The time required is 3 * 2 * 2 = 12 minutes. - The fourth mechanic will repair four cars. The time required is 1 * 4 * 4 = 16 minutes. It can be proved that the cars cannot be repaired in less than 16 minutes.
Example 2:
Input: ranks = [5,1,8], cars = 6 Output: 16 Explanation: - The first mechanic will repair one car. The time required is 5 * 1 * 1 = 5 minutes. - The second mechanic will repair four cars. The time required is 1 * 4 * 4 = 16 minutes. - The third mechanic will repair one car. The time required is 8 * 1 * 1 = 8 minutes. It can be proved that the cars cannot be repaired in less than 16 minutes.
Constraints:
1 <= ranks.length <= 1051 <= ranks[i] <= 1001 <= cars <= 106This approach leverages binary search to find the minimum time required to repair all the cars. We observe that if a mechanic with rank r repairs n cars, they can do so in r * n^2 minutes. Our goal is to minimize the maximum amount of time any mechanic would spend to repair the cars. We perform binary search on time to decide the least time possible where all cars can be finished.
The key observation here is that given a time, we can check if it's possible for all cars to be repaired within that time by distributing cars among all mechanics optimally.
In this C solution, we define a helper function canRepair that checks if it's possible to repair all cars within a given time. This function computes how many cars each mechanic can potentially repair within that time. The main function uses binary search over time to find the minimum time required to repair all cars.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n * log(cars * max_rank * cars)), where max_rank is the highest rank value.
Space Complexity: O(1).
This approach uses a greedy algorithm with a priority queue (or a min-heap) to assign cars to mechanics such that the total time taken is minimized. The priority queue is used to always pick the mechanic who will complete their next car repair in the shortest time. This approach minimizes the overall time in a different manner compared to the binary search method.
This C++ solution uses a priority queue to simulate a process where each mechanic receives cars to repair in a greedy manner. It keeps track of each mechanic's total time and cars repaired, and always picks the mechanic with the current least time to assign the next car.
Java
Python
Time Complexity: O(cars * log(n)).
Space Complexity: O(n).
| Approach | Complexity |
|---|---|
| Binary Search on Minimum Time | Time Complexity: |
| Greedy with Priority Queue | Time Complexity: |
Minimum Time to Repair Cars - Leetcode 2594 - Python • NeetCodeIO • 11,017 views views
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