How do you solve Minimum Time to Repair Cars on LeetCode?

Minimum Time to Repair Cars (LeetCode #2594) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log(max_time)) time complexity and O(1) space complexity. Key insight: Binary search can significantly reduce the time complexity.

What is the brute force solution for Minimum Time to Repair Cars?

The brute force approach for Minimum Time to Repair Cars is Brute Force, with O(n²) time complexity and O(1) space complexity. We can calculate the time taken for every possible distribution of cars among mechanics and find the maximum time required. This is simple but inefficient as it checks every combination. The optimal Optimal Solution approach improves this to O(n log(max_time)).

What algorithm or data structure is used to solve Minimum Time to Repair Cars?

Minimum Time to Repair Cars uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Greedy Algorithms.

What are the interview tips for Minimum Time to Repair Cars?

Clarify the problem requirements and constraints. Discuss your thought process and potential approaches before coding. Test your solution with edge cases during the interview.

What are common mistakes when solving Minimum Time to Repair Cars?

Not considering simultaneous work by mechanics. Overlooking the need for binary search in optimization problems.

#2594

Minimum Time to Repair Cars

Medium

ArrayBinary SearchBinary SearchGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log(max_time))
Space	O(1)	O(1)

💡

Intuition

Time O(n log(max_time))Space O(1)

We can use binary search to find the minimum time required to repair all cars. By checking if a given time can repair all cars, we can efficiently narrow down the possible minimum time.

⚙️

Algorithm

5 steps

1Step 1: Set low = 1 and high = max(ranks) * cars * cars (upper bound for time).
2Step 2: While low < high, calculate mid = (low + high) // 2.
3Step 3: Check if all cars can be repaired in 'mid' time using a helper function.
4Step 4: If yes, set high = mid; otherwise, set low = mid + 1.
5Step 5: Return low as the minimum time.

solution.py18 lines

1# Full working Python code
2
3def canRepairInTime(ranks, cars, time):
4    total_cars = 0
5    for r in ranks:
6        total_cars += int((time // r) ** 0.5)
7    return total_cars >= cars
8
9
10def minTimeToRepairCars(ranks, cars):
11    low, high = 1, max(ranks) * cars * cars
12    while low < high:
13        mid = (low + high) // 2
14        if canRepairInTime(ranks, cars, mid):
15            high = mid
16        else:
17            low = mid + 1
18    return low

ℹ

Complexity note: The time complexity is O(n log(max_time)) because we are performing a binary search over time and checking the number of cars that can be repaired in O(n) time.

1Binary search can significantly reduce the time complexity.
2Understanding how to distribute tasks among workers is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.