Efficient Car Repair Scheduling: Binary Search for Minimum Time with Rank-Based Mechanics
Intuition
Each mechanic repairs cars at a speed determined by their rank.
A mechanic with rank r needs r * n^2 minutes to repair n cars.
2The goal is to find th
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
1. Each mechanic repairs cars at a speed determined by their rank.
2.  A mechanic with rank `r` needs `r * n^2` minutes to repair `n` cars.
3. 2The goal is to find the minimum time required for all mechanics, working in parallel, to repair `cars` cars.
# Approach
<!-- Describe your approach to solving the problem. -->
1.  Use binary search on the possible time range (1 to `min(ranks) * cars^2`).
2. For each midpoint, check if all cars can be repaired using a helper function.
3. The helper function iterates through mechanics and calculates the maximum number of cars each can repair in the given time.
4. If the total is enough, decrease the search space; otherwise, increase it.
# Complexity
- Let `N` be the number of mechanics and `M` be `min(ranks) * cars^2` (upper bound on time).
- Binary search operates in O(log M) time.
- Each iteration of binary search calls `can_repaired`, which runs in O(N).
- Thus, the overall time complexity is O(N log M).
- Space complexity: O(1) since only a few extra variables are used.
# 🚀🔥 Keep coding and stay awesome!
# Code
```python3 []
class Solution:
    def repairCars(self, ranks, cars):
        """
        Finds the minimum time required to repair all cars given the mechanics' ranks.
        """
        
        # Helper function to check if a given time limit is sufficient to repair all cars
        def can_repaired(time):
            total_cars_repaired = 0
            for rank in ranks:
                total_cars_repaired += int((time / rank) ** 0.5)  # Each mechanic repairs sqrt(time / rank) cars
            return total_cars_repaired >= cars
        
        # Binary search boundaries
        left, right = 1, min(ranks) * cars * cars  # Upper bound is worst-case scenario
        
        # Perform binary search on time
        while left <= right:
            mid = (left + right) // 2  # Middle point as a potential minimum time
            if can_repaired(mid):
                right = mid - 1  # Try to find a smaller valid time
            else:
                left = mid + 1  # Increase time as current is insufficient
        
        return left  # The smallest valid time found
```
['Array', 'Hash Table', 'Binary Search', 'Divide and Conquer', 'Dynamic Programming', 'Binary Search Tree', 'Sliding Window', 'Heap (Priority Queue)', 'Prefix Sum', 'Python3']