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 <= 106The key observation in #2594 Minimum Time to Repair Cars is that each mechanic with rank r takes r * n^2 time to repair n cars. Instead of simulating repairs directly, we search for the minimum time in which all cars can be repaired.
A common strategy is to apply binary search on time. The search range represents possible repair times. For a candidate time T, we calculate how many cars each mechanic could repair within that time using the inverse of the repair formula. Summing across mechanics tells us whether the target number of cars can be repaired within T.
If the total repaired cars meet or exceed the requirement, we try a smaller time; otherwise, we increase the time. This monotonic feasibility condition makes binary search efficient.
The approach runs in O(m log(maxTime)), where m is the number of mechanics, with O(1) extra space.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Binary Search on Time | O(m log(maxTime)) | O(1) |
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Use these hints if you're stuck. Try solving on your own first.
For a predefined fixed time, can all the cars be repaired?
Try using binary search on the answer.
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Yes, problems similar to this are common in technical interviews at companies like FAANG. They test understanding of binary search on answer, mathematical modeling, and optimization techniques.
Binary search is used because the number of cars that can be repaired increases as time increases. This monotonic property allows us to efficiently search for the minimum feasible time instead of testing every possible value.
The solution primarily uses arrays to store the mechanics' ranks. Apart from simple arithmetic calculations and iteration, no advanced data structures are required, keeping the space usage minimal.
The optimal approach uses binary search on the possible time range. For each candidate time, we compute how many cars each mechanic can repair and check if the total meets the required number. This works because the feasibility condition is monotonic.