Minimum Time to Repair Cars

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Binary Search

You are given an integer array ranks representing the ranks of some mechanics. ranks_i is the rank of the i^th 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 <= 10⁵
1 <= ranks[i] <= 100
1 <= cars <= 10⁶

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the values in the `ranks` array and the value of `cars`? Specifically, what are the maximum possible values?
Can `ranks` contain zero or negative values? If so, what should the behavior be?
If `cars` is zero, should I return zero or is that an invalid input?
Is there always a solution possible? Meaning, will `ranks` always have at least one mechanic?
Are the ranks integers? Or can they be floating point numbers?

Brute Force Solution

Approach

The brute force method for this problem involves trying out every possible amount of time it could take to repair all the cars. We essentially simulate the car repairs happening and see if the given time allows all cars to be fixed.

Here's how the algorithm would work step-by-step:

Start by guessing the absolute minimum amount of time needed to fix all the cars. This is a starting point.
Then guess the next possible amount of time, incrementally increasing it.
For each guessed time, figure out how many cars each mechanic could repair within that time.
Add up the total number of cars all mechanics could repair in the given time.
If the total number of cars that *could* be repaired is greater than or equal to the number of cars that *need* to be repaired, then this time is a possible answer.
If the total number of cars is less than needed, then try a bigger amount of time.
Keep checking larger amounts of time until you find the smallest amount of time that allows all cars to be repaired.

Code Implementation

def minimum_time_to_repair_cars_brute_force(ranks, cars):
    minimum_possible_time = 0
    maximum_possible_time = max(ranks) * cars * cars

    current_time = minimum_possible_time

    while current_time <= maximum_possible_time:
        cars_repaired_at_current_time = 0

        # Check how many cars each mechanic can repair in current_time
        for rank in ranks:
            cars_repaired_at_current_time += int((current_time / rank)**0.5)

        # If we can repair enough cars, this is a valid time
        if cars_repaired_at_current_time >= cars:
            return current_time

        current_time += 1

    return maximum_possible_time

Big(O) Analysis

Time Complexity

O(n * m * log(target)) – The described brute force approach involves iterating through possible time values until a valid time is found. Determining the upper bound of the time values depends on the specific values in the ranks array and the number of cars. Let's assume the upper bound is 'target'. The algorithm effectively performs a linear search (though this is often implicitly implemented as a binary search) from the minimum possible time (likely 1) up to 'target'. For each guessed time value, it iterates through each of the 'n' mechanics in the ranks array and calculates how many cars each mechanic can repair in the given time, this takes O(n) operations, after which it sums the cars repaired, taking an additional O(n) time, resulting in a O(n) operation inside the loop. This repeats m times, where m is proportional to the range of the time values. In total this is O(n * m * log(target)) where n is the ranks array length, m is the target value described above and log(target) represent a binary search for the optimal value.

Space Complexity

O(1) – The brute force method described primarily involves iterative calculations and comparisons. No significant auxiliary data structures like lists, maps, or sets are created to store intermediate results or track visited states. The algorithm only needs to store a few variables to represent the current guessed time, the number of cars repaired by each mechanic, and the total number of repaired cars. Since the space usage is independent of the number of mechanics or cars, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The core idea is to efficiently guess the minimum time needed to repair all cars. We can use a binary search-like approach to repeatedly narrow down the possible time range, checking if a given time is sufficient by figuring out how many cars each mechanic can repair within that time.

Here's how the algorithm would work step-by-step:

First, consider the range of possible times. The lowest possible time is when the fastest mechanic repairs only one car. The highest time is when the slowest mechanic repairs all the cars.
Pick a time in the middle of this range and check if that's enough time to repair all the cars.
To check, go through each mechanic and figure out how many cars they can repair within the chosen time.
Add up the number of cars all the mechanics can repair. If it's more than or equal to the total number of cars, it means the time we guessed is high enough.
If the guessed time is high enough, try a smaller time (search in the lower half of the range). If not, try a bigger time (search in the upper half of the range).
Keep repeating this process of guessing a time, checking if it's sufficient, and narrowing the range until you find the smallest possible time that allows all the cars to be repaired.

Code Implementation

def minimum_time_to_repair_cars(ranks, number_of_cars):
    left_time = min(ranks) * 1
    right_time = max(ranks) * number_of_cars * number_of_cars

    while left_time < right_time:
        mid_time = (left_time + right_time) // 2

        cars_repaired = 0
        # Calculate the total cars that can be repaired by all mechanics.
        for rank in ranks:
            cars_repaired += (mid_time // rank)**0.5

        # Adjust the search space based on whether enough cars can be repaired.
        if cars_repaired >= number_of_cars:
            right_time = mid_time

        # Narrow search to the upper half.
        else:
            left_time = mid_time + 1

    return left_time

Big(O) Analysis

Time Complexity

O(n * log(m)) – The algorithm employs a binary search strategy where n is the length of the ranks array (number of mechanics) and m is the maximum possible time required to repair all cars. Inside the binary search loop, which runs in O(log(m)) time, we iterate through the ranks array to compute the number of cars each mechanic can repair within the guessed time. This iteration takes O(n) time. Therefore, the overall time complexity is O(n * log(m)).

Space Complexity

O(1) – The provided algorithm primarily involves a binary search-like approach to find the minimum time. It iteratively guesses a time and checks if it's sufficient, without using any auxiliary data structures that scale with the input size. Only a few constant space variables are used to store the low, high, and mid values during the binary search and to iterate through the mechanics. The number of mechanics and cars affects the computations but not the auxiliary space used, resulting in constant space complexity.

Edge Cases

Case	How to Handle
Empty ranks array.	Return 0 or throw an exception as no cars can be repaired.
Zero cars to repair.	Return 0 since no time is needed if no cars need repairing.
Single mechanic and single car.	Calculates repair time directly using the single mechanic's rank.
All mechanics have the same rank.	Binary search should converge to the correct answer regardless of identical ranks.
One mechanic has a very high rank.	Binary search space should be adjusted correctly with high repair times involved.
Large number of cars and mechanics, potential for integer overflow when calculating repair time.	Use long long to avoid overflow when calculating timetime and rank time * time.
Mechanic rank is very small or 1.	The binary search will converge to the lowest possible repair time.
Very large number of cars and large rank values causing the search space to be extremely big.	Binary Search is efficient and works as expected due to logarithmic time complexity even with a wide search space.

Empty ranks array.

How to Handle:

Return 0 or throw an exception as no cars can be repaired.

Zero cars to repair.

How to Handle:

Return 0 since no time is needed if no cars need repairing.

Single mechanic and single car.

How to Handle:

Calculates repair time directly using the single mechanic's rank.

All mechanics have the same rank.

How to Handle:

Binary search should converge to the correct answer regardless of identical ranks.

One mechanic has a very high rank.

How to Handle:

Binary search space should be adjusted correctly with high repair times involved.

Large number of cars and mechanics, potential for integer overflow when calculating repair time.

How to Handle:

Use long long to avoid overflow when calculating time*time and rank * time * time.

Mechanic rank is very small or 1.

How to Handle:

The binary search will converge to the lowest possible repair time.

Very large number of cars and large rank values causing the search space to be extremely big.

How to Handle:

Binary Search is efficient and works as expected due to logarithmic time complexity even with a wide search space.