Mathematical Derivation & Inversion of the Repair Function

If a mechanic with rank $r$ repairs $n$ cars in time $t$, their time equation is:

This formulation provides our monotonic predicate: in any allocated time $t$, mechanic $i$ can independently repair $\lfloor\sqrt{t / r_i}\rfloor$ cars. Because all mechanics work simultaneously in parallel, the total number of cars repaired across all mechanics in time $t$ is:

Monotonicity & Binary Search on Feasibility

Notice that as time $t$ increases, $\sqrt{t / r_i}$ is monotonically non-decreasing. Therefore, $\text{TotalCars}(t)$ is also monotonically non-decreasing. This enables binary search:

• If $\text{TotalCars}(t) \ge \text{cars}$, time $t$ is sufficient to repair all cars. We record $t$ as a valid answer and attempt to find a smaller time in $[L, t - 1]$.
• If $\text{TotalCars}(t) < \text{cars}$, time $t$ is insufficient. We must search in $[t + 1, R]$.

Establishing Tight Search Bounds & Overflow Safeguards

To ensure our binary search terminates swiftly and avoids 64-bit integer overflow:

• Lower Bound ($L$): $1$ minute (the minimum possible non-zero time).
• Upper Bound ($R$): What if only a single mechanic repaired all cars? The optimal candidate would be the mechanic with the minimum rank $\min(ranks)$. That single mechanic repairs all cars in $\min(ranks) \cdot \text{cars}^2$ minutes. This provides an absolute upper bound $R = \min(ranks) \cdot \text{cars}^2$.

Given $\text{cars} \le 10^6$ and $\min(ranks) \le 100$, $R \le 100 \times (10^6)^2 = 10^{14}$, which fits safely within a standard signed 64-bit integer (long long, max value $\approx 9.22 \times 10^{18}$). When multiplying minRank * cars * cars in C++, using 1LL * minRank * cars * cars explicitly prevents intermediate 32-bit integer overflow.

Numerical Stability: Integer Square Root vs. Floating-Point Precision

In high-precision competitive programming and numerical systems, using standard 64-bit floating-point square root (std::sqrt) on values up to $10^{14}$ is mathematically safe because IEEE 754 double-precision floats provide 53 bits of mantissa precision (exact integer representation up to $2^{53} - 1 \approx 9.007 \times 10^{15}$). Since our maximum time is $10^{14}$, double-precision casting introduces zero truncation or rounding errors.

However, in systems where inputs can exceed $2^{53}$, reliance on hardware floating-point units can introduce off-by-one errors. In such scenarios, an integer square root algorithm (using integer binary search or integer Newton-Raphson) guarantees exact results:

// Safe 64-bit integer square root without floating-point conversion
long long integer_sqrt(long long n) {
    if (n <= 0) return 0;
    long long x0 = n / 2;
    if (x0 == 0) return 1;
    long long x1 = (x0 + n / x0) / 2;
    while (x1 < x0) {
        x0 = x1;
        x1 = (x0 + n / x0) / 2;
    }
    return x0;
}

Optimized C++20 Implementation with Frequency Map

In the standard approach, checking feasibility requires iterating over all $N$ mechanics ($N \le 10^5$), giving $O(N \log R)$ time. However, notice that rank values are bounded: $1 \le ranks[i] \le 100$. By counting mechanic frequencies, we reduce the verification loop from $10^5$ to at most 100 iterations!

#include <vector>
#include <cmath>
#include <algorithm>
#include <iostream>

class Solution {
public:
    long long repairCars(std::vector<int>& ranks, int cars) {
        // Frequency array: ranks are bounded between 1 and 100
        int count[101] = {0};
        int minRank = 101;
        for (int r : ranks) {
            count[r]++;
            if (r < minRank) minRank = r;
        }

        // Search space bounds:
        // L = 1, R = minRank * cars^2
        long long left = 1;
        long long right = 1LL * minRank * cars * cars;
        long long result = right;

        // Monotonic feasibility check
        auto canRepairInTime = [&](long long time) -> bool {
            long long totalRepaired = 0;
            for (int r = 1; r <= 100; ++r) {
                if (count[r] == 0) continue;
                // Cars repaired by all mechanics of rank r in 'time' minutes
                totalRepaired += 1LL * count[r] * static_cast<long long>(std::sqrt(time / r));
                if (totalRepaired >= cars) return true; // Early exit optimization
            }
            return totalRepaired >= cars;
        };

        while (left <= right) {
            long long mid = left + (right - left) / 2;
            if (canRepairInTime(mid)) {
                result = mid;       // Valid time found; try to minimize further
                right = mid - 1;
            } else {
                left = mid + 1;     // Insufficient time; increase lower bound
            }
        }

        return result;
    }
};

int main() {
    Solution solver;
    std::vector<int> ranks = {4, 2, 3, 1};
    int cars = 10;
    
    long long minTime = solver.repairCars(ranks, cars);
    std::cout << "Minimum time to repair " << cars << " cars: " << minTime << " minutes\n";
    // Expected output: 16 minutes
    return 0;
}

Distributed Systems Analogy: Heterogeneous Cloud Worker Scheduling

The mathematical model in LeetCode 2594 directly parallels real-world autoscaling architectures in cloud engineering. Consider a distributed queuing cluster (e.g. RabbitMQ or Kafka) where heterogeneous worker nodes pull tasks from a shared backlog:

• High-performance GPU instances process batches at near-linear latency (low rank $r$).
• Burstable CPU instances suffer quadratic latency degradation under heavy load due to thermal throttling and thread context switching (higher rank $r$).
• Determining the minimum completion deadline across the cluster requires inverting worker capacity functions and solving for parallel throughput using binary search predicates.

Complexity Analysis

With $K \le 100$ and $\log_2(10^{14}) \approx 47$ iterations, the frequency-optimized feasibility check completes in less than 4,700 operations—executing in under 1 millisecond.