If you’ve ever wondered how to efficiently manage a garage with multiple mechanics repairing cars, this problem might feel familiar. In this blog, we’ll explore a classic computational problem known as “Minimum Time to Repair Cars,” which involves determining the minimum time needed to repair all cars with a set of mechanics, each having different repair speeds. We’ll break down the problem, explain a brute force approach, and highlight why this approach may not be efficient for larger inputs.

You are given a list of mechanics, each with a specific rank. The rank of a mechanic determines how fast they can repair cars: the time taken to repair n cars by a mechanic with rank r is calculated as r * n^2 minutes.

Given an array of mechanics’ ranks and the total number of cars that need repair, your goal is to minimize the maximum time any mechanic spends repairing their assigned cars. Importantly, all mechanics can repair cars simultaneously.

Example 1:

• Input: ranks = [4, 2, 3, 1], cars = 10
• Output: 16 minutes

The goal is to figure out how to allocate the cars so that the repair time for the slowest mechanic (the one with the highest repair time) is minimized.

Example 2:

• Input: ranks = [5, 1, 8], cars = 6

Brute Force Approach to Solving the Problem

In order to solve the “Minimum Time to Repair Cars” problem, a brute force approach would involve trying all possible ways to assign cars to the mechanics and then determining the time it would take for each possible distribution. The key idea is to minimize the maximum time any mechanic will need.

Steps for a Brute Force Solution:

1. Try All Possible Assignments of Cars to Mechanics:
   • You start by distributing the cars among the mechanics. For example, if you have 10 cars and 4 mechanics, you could try assigning 2 cars to two mechanics and 3 cars to two others. There are countless possible ways to distribute these cars.
2. Calculate Time for Each Assignment:
   • For each distribution of cars, you calculate how long each mechanic will take. If a mechanic with rank r[i] repairs n cars, the time taken is calculated as r[i] * n^2.
3. Find the Maximum Time:
   • After calculating the time for all mechanics in a specific assignment, you find the mechanic with the longest repair time. This value is the maximum time for that particular assignment.
4. Select the Minimum of the Maximum Times:
   • You repeat the above steps for all possible assignments, and then pick the one with the minimum maximum time. This would be the most optimal distribution of cars among mechanics.

Example Walkthrough:

Let’s take a look at how the brute force approach would work with an example.

Given:

• Mechanics’ ranks: [4, 2, 3, 1]
• Total cars: 10

One possible way to assign the cars is:

• Mechanic 1 (rank 4) repairs 2 cars: This takes 4 * 2^2 = 16 minutes.
• Mechanic 2 (rank 2) repairs 2 cars: This takes 2 * 2^2 = 8 minutes.
• Mechanic 3 (rank 3) repairs 2 cars: This takes 3 * 2^2 = 12 minutes.
• Mechanic 4 (rank 1) repairs 4 cars: This takes 1 * 4^2 = 16 minutes.

Here, the maximum time taken is 16 minutes. By trying all possible distributions, you can determine the minimal time.

Optimization Strategy: Binary Search + Greedy Approach

We can optimize the solution using a binary search combined with a greedy approach. The idea is to search for the minimum maximum time in which all cars can be repaired.

How It Works:

1. Binary Search:
   • Since the minimum possible time to repair all cars is the time taken by the fastest mechanic to repair one car, and the maximum time is the time taken by the slowest mechanic to repair all cars, we can apply binary search on the time.
2. Greedy Check:
   • For each time in the binary search range, we use a greedy approach to check if it’s feasible to repair all cars within that time. We try to assign cars to mechanics in such a way that no mechanic exceeds the current time being tested.

Step-by-Step Explanation:

1. Binary Search on Time:

• The binary search will focus on finding the minimum possible time. We will search between a lower bound and an upper bound:
  • Lower Bound: The fastest possible time, which would be the time for the fastest mechanic to repair just one car. This is min(ranks) * 1^2 = min(ranks).
  • Upper Bound: The slowest possible time, which would be the time for the slowest mechanic to repair all cars. This is max(ranks) * cars^2.
• We perform binary search within this range to find the smallest possible maximum repair time.

2. Greedy Check for Feasibility:

• For each time value T in the binary search, we will check if it’s possible to assign cars to mechanics such that no mechanic exceeds T minutes.
• The greedy strategy works as follows:
  • For each mechanic, we calculate how many cars they can repair without exceeding the time T. Specifically, we solve the equation r * n^2 <= T, where r is the mechanic’s rank and n is the number of cars they can repair. This can be rearranged to find n <= sqrt(T / r).
  • Sum the number of cars that all mechanics can repair in time T. If the total number of cars is at least equal to the required number of cars, then it’s feasible to repair all cars in time T.
• If it’s feasible to repair all cars in time T, we reduce the upper bound of our binary search to search for a smaller time. If it’s not feasible, we increase the lower bound.

3. Edge Cases:

• One mechanic: If there’s only one mechanic, they will have to repair all the cars, and the answer will simply be rank[0] * cars^2.
• One car: If there is only one car, the answer is simply min(ranks), as the fastest mechanic will repair it.

Code Implementation of the Optimized Solution:

Here’s how you can implement the binary search and greedy approach to solve the problem efficiently:

import math

def repairCars(ranks, cars):
    # Binary search bounds
    left, right = min(ranks), max(ranks) * cars * cars

    def canRepairInTime(maxTime):
        # Check if it's possible to repair all cars within maxTime
        totalCarsRepaired = 0
        for rank in ranks:
            totalCarsRepaired += int(math.sqrt(maxTime / rank))
            if totalCarsRepaired >= cars:
                return True
        return totalCarsRepaired >= cars

    # Perform binary search
    while left < right:
        mid = (left + right) // 2
        if canRepairInTime(mid):
            right = mid  # Try for smaller max time
        else:
            left = mid + 1  # Increase the time

    return left

Explanation of the Code:

1. Binary Search Bounds:
   • left is the minimum time possible, which is the fastest mechanic repairing just one car: min(ranks).
   • right is the maximum time possible, which is the slowest mechanic repairing all the cars: max(ranks) * cars^2.
2. Greedy Check (canRepairInTime function):
   • For a given time maxTime, we calculate the total number of cars each mechanic can repair without exceeding this time. The number of cars a mechanic with rank r can repair is sqrt(maxTime / r).
   • We check if the total number of cars repaired by all mechanics is greater than or equal to the total number of cars needed.
3. Binary Search Logic:
   • We continue to adjust the search bounds (left and right) based on whether the current mid-time is feasible or not. If it’s feasible, we reduce the time by adjusting right. If it’s not feasible, we increase the time by adjusting left.

Time Complexity:

• Binary Search: The binary search will run for about O(log(maxTime)) iterations, where maxTime is the maximum possible time.
• Greedy Check: For each binary search iteration, we iterate over all mechanics to check if the distribution is feasible. This takes O(n) time, where n is the number of mechanics.

Thus, the overall time complexity is O(n log(maxTime)), where n is the number of mechanics, and maxTime is the maximum time any mechanic would take to repair all cars.

This is much more efficient than the brute force approach, which could potentially have an exponential time complexity due to the combinatorial nature of the car distributions.