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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2501_2600.s2594_minimum_time_to_repair_cars;

// #Medium #Array #Binary_Search #2023_08_23_Time_15_ms_(97.28%)_Space_53.8_MB_(85.07%)

/**
 * 2594 - Minimum Time to Repair Cars\.
 *
 * Medium
 *
 * You are given an integer array `ranks` representing the **ranks** of some mechanics. ranks_i is the rank of the i^th mechanic. A mechanic with a rank `r` can repair n cars in r * n² minutes.
 *
 * You are also given an integer `cars` representing the total number of cars waiting in the garage to be repaired.
 *
 * Return _the **minimum** time taken to repair all the cars._
 *
 * **Note:** All the mechanics can repair the cars simultaneously.
 *
 * **Example 1:**
 *
 * **Input:** ranks = [4,2,3,1], cars = 10
 *
 * **Output:** 16
 *
 * **Explanation:**
 *
 * - The first mechanic will repair two cars. The time required is 4 \* 2 \* 2 = 16 minutes.
 *
 * - The second mechanic will repair two cars. The time required is 2 \* 2 \* 2 = 8 minutes.
 *
 * - The third mechanic will repair two cars. The time required is 3 \* 2 \* 2 = 12 minutes.
 *
 * - The fourth mechanic will repair four cars. The time required is 1 \* 4 \* 4 = 16 minutes.
 *
 * It can be proved that the cars cannot be repaired in less than 16 minutes.
 *
 * **Example 2:**
 *
 * **Input:** ranks = [5,1,8], cars = 6
 *
 * **Output:** 16
 *
 * **Explanation:**
 *
 * - The first mechanic will repair one car. The time required is 5 \* 1 \* 1 = 5 minutes.
 *
 * - The second mechanic will repair four cars. The time required is 1 \* 4 \* 4 = 16 minutes.
 *
 * - The third mechanic will repair one car. The time required is 8 \* 1 \* 1 = 8 minutes.
 *
 * It can be proved that the cars cannot be repaired in less than 16 minutes.
 *
 * **Constraints:**
 *
 * *   1 <= ranks.length <= 10⁵
 * *   `1 <= ranks[i] <= 100`
 * *   1 <= cars <= 10⁶
**/
public class Solution {
    public long repairCars(int[] ranks, int cars) {
        long low = 0;
        long high = 100000000000000L;
        long pivot;
        while (low <= high) {
            pivot = low + (high - low) / 2;
            if (canRepairAllCars(ranks, cars, pivot)) {
                high = pivot - 1;
            } else {
                low = pivot + 1;
            }
        }
        return low;
    }

    private boolean canRepairAllCars(int[] ranks, int cars, long totalMinutes) {
        int repairedCars = 0;
        for (int i = 0; i < ranks.length && repairedCars < cars; i++) {
            repairedCars += (int) Math.sqrt((double) totalMinutes / ranks[i]);
        }
        return repairedCars >= cars;
    }
}