Quick Summary of LeetCode 2594
The LeetCode 2594: Minimum Time to Repair Cars problem asks us to find the absolute minimum time required for a group of mechanics to repair a specified number of cars. Each mechanic is given a rank, and a mechanic with rank r repairs n cars in r × n² minutes. The most optimal approach to solving this involves using a binary search on the answer space, achieving an efficient time complexity of O(R log(Tmax)), where R is the number of mechanics. Below is the detailed C++ solution and algorithmic breakdown.
Algorithmic Approach: Binary Search on Answer Space
Because the capacity of mechanics to repair cars increases monotonically with time, we can binary search the answer space (time). This monotonic property means if a set of cars can be repaired in T minutes, they can also be repaired in T + 1 minutes. Thus, our goal is to find the minimum valid T.
The minimum possible time is 1 minute, and the maximum upper bound is the time it takes the fastest mechanic to repair all the cars alone. This upper bound is evaluated as: ranks[0] × cars². Using binary search, we continually halve our search space:
- For a given midpoint time T, a mechanic with rank r can repair a maximum of n = ⌊√(T / r)⌋ cars.
- We calculate the total sum of cars repaired by all mechanics within time T.
- If the total sum is greater than or equal to the required number of cars, the time T is valid, and we attempt to search for an even smaller time. Otherwise, we must increase T.
C++ Implementation
Here is the robust C++ implementation to solve LeetCode 2594 using binary search. This code minimizes the search space effectively while carefully avoiding integer overflow by leveraging the long long data type.
#include <vector>
#include <cmath>
using namespace std;
class Solution {
public:
long long repairCars(vector<int>& ranks, int cars) {
long long left = 1;
long long right = ranks[0] * (long long)cars * cars;
long long result = right; // Initialize with upper bound
auto countRepairCars = [&ranks](long long time) {
long long repaired = 0;
for (int r : ranks) {
repaired += (long long)sqrt((double)time / r);
}
return repaired;
};
while (left <= right) {
long long midpoint = left + (right - left) / 2;
long long repaired = countRepairCars(midpoint);
if (repaired >= cars) {
result = midpoint; // Found a valid time, try to minimize it
right = midpoint - 1;
} else {
left = midpoint + 1;
}
}
return result;
}
};
Complexity Analysis
- Time Complexity: O(R log(Tmax)), where R is the size of the ranks array and Tmax = ranks[0] × cars². This performance scales excellently even for large inputs, ensuring execution within standard competitive programming limits.
- Space Complexity: O(1) constant space. The memory overhead is strictly limited to auxiliary variables, independent of the input size.
Frequently Asked Questions (FAQ)
Why do we use binary search instead of dynamic programming?
Dynamic programming or greedy assignment of cars one by one would be too slow because the number of cars can be exceptionally large (up to 1,000,000). Binary search efficiently zeroes in on the exact minimum time required across the vast potential answer space in logarithmic steps.
What happens if multiple mechanics have the same rank?
The algorithm effortlessly handles multiple mechanics with identical ranks. The countRepairCars function loops through all available mechanics independently, accumulating their capacity within the given midpoint time. Therefore, duplicate ranks simply contribute proportionally to the total repaired car count.
How does the algorithm prevent integer overflow?
Because the maximum time Tmax can grow massively (e.g., a rank of 100 with 1,000,000 cars results in 100 × 1,000,000²), standard 32-bit integers will overflow. We use 64-bit integers (long long in C++) for the left, right, midpoint, and result variables to ensure complete mathematical precision and safety.
Can this algorithm be optimized further?
Since the array of ranks can have up to 100,000 elements, you can pre-process the ranks to count the frequency of each rank (from 1 to 100). Iterating through unique ranks instead of the entire array reduces the loop size in countRepairCars, further optimizing the time complexity.