2594. Minimum Time to Repair Cars

Intuition

This problem requires us to find the minimum time needed for mechanics with different ranks to repair a certain number of cars. A key insight is understanding that the time required for a mechanic to repair cars follows a formula: a mechanic with rank r takes r * n² time to repair n cars.

Conversely, in a given amount of time t, a mechanic with rank r can repair ⌊√(t/r)⌋ cars. Since we want to find the minimum time for all mechanics to collectively repair all cars, we can use binary search on the time to efficiently find this minimum value.

Approach

1. Define a search space for the minimum time:

   • Lower bound (low): 1 (minimum possible time)
   • Upper bound (high): max(ranks) * cars * cars (worst-case scenario where the slowest mechanic repairs all cars)

2. Create a helper function canBeRepairedAll(time) that determines if all cars can be repaired within a specific time:

   • For each mechanic with rank r, calculate how many cars they can repair in time t: ⌊√(t/r)⌋
   • Sum up the number of cars all mechanics can repair
   • Return true if this sum is greater than or equal to the total number of cars

3. Perform binary search on the time:

   • If canBeRepairedAll(mid) returns true, try a smaller time (high = mid - 1)
   • Otherwise, increase the time (low = mid + 1)

4. Return low as the minimum time required when the binary search completes

Code

function repairCars(ranks: number[], cars: number): number {
    // Initialize the binary search bounds
    let low = 1;
    // Calculate the upper bound for binary search
    // The maximum time is the time required to repair the most time-consuming car 'cars' times
    let high = Math.max(...ranks) * cars * cars;
 
    // Helper function to check if all cars can be repaired within a given time
    const canBeRepairedAll = (time: number): boolean => {
        let maxCarsRepaired = ranks.reduce((total, rank) => total + Math.floor(Math.sqrt(time / rank)), 0);
        return maxCarsRepaired >= cars;
    };
 
    // Binary search to find the minimum time required
    while (low <= high) {
        let mid = Math.floor(low + (high - low) / 2);
        if (canBeRepairedAll(mid)) {
            high = mid - 1; // Try to find a smaller valid time
        } else {
            low = mid + 1; // Increase the time if not all cars can be repaired
        }
    }
 
    // The minimum time required to repair all cars
    return low;
};

Time Complexity

• Time Complexity: O(n log(m * k))

  • n: number of mechanics (length of ranks array)
  • m: maximum rank value
  • k: number of cars
  • Binary search takes O(log(m * k)) iterations
  • In each iteration, we check all mechanics, which takes O(n)

• Space Complexity: O(1)

  • We only use a constant amount of extra space regardless of input size

The binary search approach is efficient because it narrows down the search space logarithmically, allowing us to quickly find the minimum time required for all mechanics to repair all cars.