Solution #2594 - Mridul/Edited - 16/03/2025

Daily Questions/#2594 - Minimum Time to Repair Cars - Medium/Explanation.md

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+# Explanation of the `repairCars` Code
+
+This Java class provides an efficient solution for calculating the minimum time required to repair a given number of cars, given the repair capabilities of different mechanics. The code uses a **binary search algorithm** to minimize the time.
+
+---
+
+## Class: `Solution`
+
+The `Solution` class consists of two primary methods:
+- `repairCars(int[] ranks, int cars)`
+- `numCarsFixed(int[] ranks, long minutes)`
+
+---
+
+### 1. Method: `repairCars`
+
+#### **Purpose**
+This method calculates the minimum time required to repair all the cars.
+
+#### **Parameters**
+- `ranks`: An array of integers, where each element represents the "rank" of a mechanic. A lower rank indicates a faster mechanic.
+- `cars`: An integer indicating the number of cars that need to be repaired.
+
+#### **Logic**
+1. **Initialization**:
+   - `l` is initialized to `0` (minimum possible time).
+   - `r` is initialized to the maximum possible time:
+
+
+\[
+     r = \text{(minimum rank)} \times \text{cars}^2
+     \]
+
+
+   - This is because the worst-case scenario occurs when the fastest mechanic fixes all cars, and time increases quadratically with the number of cars.
+
+2. **Binary Search**:
+   - While `l` is less than `r`, calculate the middle point `m = (l + r) / 2`.
+   - Check if it is possible to repair the required number of cars within `m` minutes by calling `numCarsFixed(ranks, m)`:
+     - If **true**, reduce the upper bound `r` to `m` (try to minimize time
+
+     Time COmplexity: O(n)
+     Space Complexity: O(n)

Daily Questions/#2594 - Minimum Time to Repair Cars - Medium/Solution.java

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+class Solution {
+    public long repairCars(int[] ranks, int cars) {
+      long l = 0;
+      long r = (long) Arrays.stream(ranks).min().getAsInt() * cars * cars;
+
+      while (l < r) {
+        final long m = (l + r) / 2;
+        if (numCarsFixed(ranks, m) >= cars)
+          r = m;
+        else
+          l = m + 1;
+      }
+
+      return l;
+    }
+
+    private long numCarsFixed(int[] ranks, long minutes) {
+      long carsFixed = 0;
+      for (final int rank : ranks)
+        carsFixed += Math.sqrt(minutes / rank);
+      return carsFixed;
+    }
+  }