Problem Description

You have an array ranks where each element represents the rank of a mechanic. A mechanic with rank r needs r * n² minutes to repair n cars.

Given:

• ranks: an integer array where ranks[i] is the rank of the i-th mechanic
• cars: the total number of cars that need to be repaired

The key points are:

• Each mechanic works independently and simultaneously
• A mechanic with rank r takes r * n² minutes to repair exactly n cars
• For example, a mechanic with rank 3 would take:
  • 3 × 1² = 3 minutes to repair 1 car
  • 3 × 2² = 12 minutes to repair 2 cars
  • 3 × 3² = 27 minutes to repair 3 cars

You need to find the minimum time required to repair all cars when all mechanics work at the same time. Each mechanic can repair any number of cars (up to the total needed), and you want to distribute the work optimally to minimize the total time.

The solution uses binary search on the time. For a given time t, we can calculate how many cars each mechanic can repair: if a mechanic has rank r, they can repair ⌊√(t/r)⌋ cars in time t. We search for the minimum time where the total number of cars that can be repaired is at least equal to the required number of cars.

Intuition

The key insight is recognizing that this is a monotonic problem - as we give mechanics more time, they can repair more cars. This monotonic property makes binary search a natural fit.

Let's think about it differently: instead of asking "how should we distribute cars to minimize time?", we can ask "given a fixed amount of time t, can we repair all the cars?" This transforms the optimization problem into a decision problem.

For any given time t, we can calculate exactly how many cars each mechanic can complete. Since a mechanic with rank r needs r * n² minutes to repair n cars, we can rearrange this formula: if we have t minutes available, then t = r * n², which means n = √(t/r). So each mechanic can repair ⌊√(t/r)⌋ cars in time t.

The beautiful part is that we don't need to figure out the optimal distribution of cars among mechanics. We just need to check if the total capacity (sum of all cars that can be repaired by all mechanics in time t) meets our requirement. If it does, we might be able to do it faster (try a smaller time). If it doesn't, we definitely need more time.

This naturally leads to binary search on the time:

• If mechanics can repair at least cars vehicles in time t, try a smaller time
• If they can't repair enough in time t, we need more time

The search space for time ranges from 0 to some upper bound. The worst case would be if we only had one mechanic with the highest rank repairing all cars sequentially, giving us an upper bound of ranks[0] * cars * cars.

Learn more about Binary Search patterns.

Solution Approach

The implementation uses binary search with a helper function to check feasibility at each time point.

Binary Search Setup:

• We use bisect_left to find the leftmost (minimum) time where the condition is satisfied
• The search range is range(ranks[0] * cars * cars), which creates values from 0 to ranks[0] * cars * cars - 1
• The upper bound ranks[0] * cars * cars represents the worst-case scenario where the first mechanic (potentially with highest rank) repairs all cars alone

Check Function: The helper function check(t) determines if all cars can be repaired within time t:

• For each mechanic with rank r, calculate how many cars they can repair: ⌊√(t/r)⌋
• We use int(sqrt(t // r)) to compute this value
  • t // r performs integer division
  • sqrt() calculates the square root
  • int() floors the result to get whole cars
• Sum up all cars that can be repaired by all mechanics
• Return True if the total is at least cars, otherwise False

How Binary Search Works Here:

1. bisect_left searches for the first position where check(t) returns True
2. It uses the key parameter to transform each time value through the check function
3. When check(t) returns False, it means we need more time
4. When check(t) returns True, it means this time is sufficient, but we try to find a smaller one
5. The algorithm converges to the minimum time where exactly cars or more can be repaired

Time Complexity:

• Binary search runs in O(log(ranks[0] * cars²)) iterations
• Each check function iterates through all mechanics: O(len(ranks))
• Overall: O(len(ranks) * log(ranks[0] * cars²))

Example Walkthrough

Let's walk through a concrete example with ranks = [4, 2, 3, 1] and cars = 10.

Step 1: Set up binary search range

• Lower bound: 0
• Upper bound: 4 × 10 × 10 = 400 (using first mechanic's rank)
• We'll search for the minimum time in range [0, 400)

Step 2: Binary search process

Let's trace through a few iterations:

Check time = 200 (midpoint)

• Mechanic rank 4: Can repair ⌊√(200/4)⌋ = ⌊√50⌋ = 7 cars
• Mechanic rank 2: Can repair ⌊√(200/2)⌋ = ⌊√100⌋ = 10 cars
• Mechanic rank 3: Can repair ⌊√(200/3)⌋ = ⌊√66.67⌋ = 8 cars
• Mechanic rank 1: Can repair ⌊√(200/1)⌋ = ⌊√200⌋ = 14 cars
• Total: 7 + 10 + 8 + 14 = 39 cars ✓ (more than 10 needed)
• Since we can repair all cars, try a smaller time

Check time = 100

• Mechanic rank 4: Can repair ⌊√(100/4)⌋ = ⌊√25⌋ = 5 cars
• Mechanic rank 2: Can repair ⌊√(100/2)⌋ = ⌊√50⌋ = 7 cars
• Mechanic rank 3: Can repair ⌊√(100/3)⌋ = ⌊√33.33⌋ = 5 cars
• Mechanic rank 1: Can repair ⌊√(100/1)⌋ = ⌊√100⌋ = 10 cars
• Total: 5 + 7 + 5 + 10 = 27 cars ✓ (still more than 10)
• Try even smaller time

Check time = 50

• Mechanic rank 4: Can repair ⌊√(50/4)⌋ = ⌊√12.5⌋ = 3 cars
• Mechanic rank 2: Can repair ⌊√(50/2)⌋ = ⌊√25⌋ = 5 cars
• Mechanic rank 3: Can repair ⌊√(50/3)⌋ = ⌊√16.67⌋ = 4 cars
• Mechanic rank 1: Can repair ⌊√(50/1)⌋ = ⌊√50⌋ = 7 cars
• Total: 3 + 5 + 4 + 7 = 19 cars ✓ (still sufficient)

Check time = 25

• Mechanic rank 4: Can repair ⌊√(25/4)⌋ = ⌊√6.25⌋ = 2 cars
• Mechanic rank 2: Can repair ⌊√(25/2)⌋ = ⌊√12.5⌋ = 3 cars
• Mechanic rank 3: Can repair ⌊√(25/3)⌋ = ⌊√8.33⌋ = 2 cars
• Mechanic rank 1: Can repair ⌊√(25/1)⌋ = ⌊√25⌋ = 5 cars
• Total: 2 + 3 + 2 + 5 = 12 cars ✓ (just enough!)

Check time = 12

• Mechanic rank 4: Can repair ⌊√(12/4)⌋ = ⌊√3⌋ = 1 car
• Mechanic rank 2: Can repair ⌊√(12/2)⌋ = ⌊√6⌋ = 2 cars
• Mechanic rank 3: Can repair ⌊√(12/3)⌋ = ⌊√4⌋ = 2 cars
• Mechanic rank 1: Can repair ⌊√(12/1)⌋ = ⌊√12⌋ = 3 cars
• Total: 1 + 2 + 2 + 3 = 8 cars ✗ (not enough)
• Need more time

Check time = 16

• Mechanic rank 4: Can repair ⌊√(16/4)⌋ = ⌊√4⌋ = 2 cars
• Mechanic rank 2: Can repair ⌊√(16/2)⌋ = ⌊√8⌋ = 2 cars
• Mechanic rank 3: Can repair ⌊√(16/3)⌋ = ⌊√5.33⌋ = 2 cars
• Mechanic rank 1: Can repair ⌊√(16/1)⌋ = ⌊√16⌋ = 4 cars
• Total: 2 + 2 + 2 + 4 = 10 cars ✓ (exactly enough!)

The binary search converges to 16 minutes as the minimum time needed to repair all 10 cars.

Verification: With 16 minutes, the optimal distribution would be:

• Mechanic rank 1 repairs 4 cars (takes 1×4² = 16 minutes)
• Mechanic rank 2 repairs 2 cars (takes 2×2² = 8 minutes)
• Mechanic rank 3 repairs 2 cars (takes 3×2² = 12 minutes)
• Mechanic rank 4 repairs 2 cars (takes 4×2² = 16 minutes)
• Total: 4 + 2 + 2 + 2 = 10 cars ✓

Solution Implementation

Time and Space Complexity

The time complexity is O(n × log M), where:

• n is the length of the ranks list (number of mechanics)
• M is the upper bound of the binary search, which is ranks[0] * cars * cars

The binary search performed by bisect_left runs in O(log M) time. For each iteration of the binary search, the check function is called, which iterates through all n elements in ranks to compute the sum, taking O(n) time. Therefore, the overall time complexity is O(n × log M).

The space complexity is O(1) as the algorithm only uses a constant amount of extra space. The check function uses a generator expression that doesn't create an intermediate list, and only maintains variables for the sum calculation and the binary search bounds.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow in Square Root Calculation

When calculating sqrt(time_limit // rank), if time_limit is very large and rank is very small, the division time_limit // rank might overflow or cause precision issues. Additionally, using integer division (//) before taking the square root can lead to incorrect results.

Problem Example:

# Incorrect approach
cars_by_this_mechanic = int(sqrt(time_limit // rank))

If time_limit = 10 and rank = 3, then 10 // 3 = 3, and sqrt(3) ≈ 1.73, giving us 1 car. However, with time 10 and rank 3, we should be able to repair more cars since 3 * 1² = 3 and 3 * 2² = 12, so only 1 car is correct but the calculation path could be misleading with larger numbers.

Solution: Use floating-point division to maintain precision:

cars_by_this_mechanic = int(sqrt(time_limit / rank))

2. Incorrect Binary Search Bounds

The upper bound calculation uses min(ranks) * cars * cars, but this assumes the most efficient mechanic repairs all cars. If you accidentally use max(ranks) or ranks[0] without sorting, you might set the bound too low or inconsistently.

Problem Example:

# Incorrect - using first element without considering it might not be minimum
max_time = ranks[0] * cars * cars

Solution: Always use the minimum rank for the upper bound or a sufficiently large value:

min_rank = min(ranks)
max_time = min_rank * cars * cars

3. Off-by-One Error in Binary Search Range

Using range(max_time) instead of range(max_time + 1) would exclude the maximum time value from the search space, potentially missing the correct answer if the minimum time equals max_time.

Problem Example:

# Incorrect - excludes max_time from search
minimum_time = bisect_left(range(max_time), True, key=can_repair_all_cars)

Solution: Include the upper bound in the search range:

minimum_time = bisect_left(range(max_time + 1), True, key=can_repair_all_cars)

4. Early Termination in Check Function

Adding an optimization to break early when enough cars are repaired might seem efficient but could lead to issues if you need to track exact distributions later.

Problem Example:

def can_repair_all_cars(time_limit: int) -> bool:
    total_cars_repaired = 0
    for rank in ranks:
        cars_by_this_mechanic = int(sqrt(time_limit / rank))
        total_cars_repaired += cars_by_this_mechanic
        if total_cars_repaired >= cars:  # Early termination
            return True  # Might miss calculating exact distribution
    return False

Solution: Keep the full calculation unless performance is critical and you're certain about not needing complete information:

def can_repair_all_cars(time_limit: int) -> bool:
    total_cars_repaired = sum(int(sqrt(time_limit / rank)) for rank in ranks)
    return total_cars_repaired >= cars