Introduction

In this blog post, we will explore how to solve the LeetCode problem "2594" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 <= 105 1 <= ranks[i] <= 100 1 <= cars <= 106

Explanation

Here's the approach, complexity analysis, and Python code to solve the problem:

• Binary Search: Perform binary search on the time range to find the minimum time required to repair all cars.
• Feasibility Check: For a given time, determine if all the cars can be repaired by all mechanics within that time.

• Adjust Search Range: Based on the feasibility check, adjust the binary search range (lower/upper bounds).

• Runtime Complexity: O(n log(m) log(cars^2 max(ranks))), where n is the number of mechanics, m is the number of cars, and max(ranks) is the maximum rank among the mechanics. This stems from the binary search and the linear check for feasibility in each iteration. Storage Complexity: O(1).

Code

    def repairCars(ranks, cars):
    """
    Calculates the minimum time to repair all cars given the ranks of mechanics.

    Args:
        ranks (list[int]): List of ranks of mechanics.
        cars (int): Total number of cars to be repaired.

    Returns:
        int: Minimum time taken to repair all cars.
    """

    def can_repair(time):
        """
        Checks if all cars can be repaired within the given time.

        Args:
            time (int): The maximum time allowed for repairs.

        Returns:
            bool: True if all cars can be repaired within the time, False otherwise.
        """
        repaired_cars = 0
        for rank in ranks:
            repaired_cars += int((time / rank)**0.5)
        return repaired_cars >= cars

    low = 0
    high = max(ranks) * cars * cars  # Max possible time
    ans = high

    while low <= high:
        mid = (low + high) // 2
        if can_repair(mid):
            ans = mid
            high = mid - 1
        else:
            low = mid + 1

    return ans