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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2101_2200.s2187_minimum_time_to_complete_trips;

// #Medium #Array #Binary_Search #2022_06_02_Time_187_ms_(95.03%)_Space_92.6_MB_(9.55%)

/**
 * 2187 - Minimum Time to Complete Trips\.
 *
 * Medium
 *
 * You are given an array `time` where `time[i]` denotes the time taken by the i^th bus to complete **one trip**.
 *
 * Each bus can make multiple trips **successively**; that is, the next trip can start **immediately after** completing the current trip. Also, each bus operates **independently**; that is, the trips of one bus do not influence the trips of any other bus.
 *
 * You are also given an integer `totalTrips`, which denotes the number of trips all buses should make **in total**. Return _the **minimum time** required for all buses to complete **at least**_ `totalTrips` _trips_.
 *
 * **Example 1:**
 *
 * **Input:** time = [1,2,3], totalTrips = 5
 *
 * **Output:** 3
 *
 * **Explanation:**
 *
 * - At time t = 1, the number of trips completed by each bus are [1,0,0].
 *
 * The total number of trips completed is 1 + 0 + 0 = 1.
 *
 * - At time t = 2, the number of trips completed by each bus are [2,1,0].
 *
 * The total number of trips completed is 2 + 1 + 0 = 3.
 *
 * - At time t = 3, the number of trips completed by each bus are [3,1,1].
 *
 * The total number of trips completed is 3 + 1 + 1 = 5.
 *
 * So the minimum time needed for all buses to complete at least 5 trips is 3. 
 *
 * **Example 2:**
 *
 * **Input:** time = [2], totalTrips = 1
 *
 * **Output:** 2
 *
 * **Explanation:**
 *
 * There is only one bus, and it will complete its first trip at t = 2.
 *
 * So the minimum time needed to complete 1 trip is 2. 
 *
 * **Constraints:**
 *
 * *   1 <= time.length <= 10⁵
 * *   1 <= time[i], totalTrips <= 10⁷
**/
public class Solution {
    public long minimumTime(int[] time, int totalTrips) {
        return bs(0, Long.MAX_VALUE, time, totalTrips);
    }

    private long bs(long left, long right, int[] time, long totalTrips) {
        if (left > right) {
            return Long.MAX_VALUE;
        }
        long mid = (left + right) >> 1;
        return isPossible(time, mid, totalTrips)
                ? Math.min(mid, bs(left, mid - 1, time, totalTrips))
                : bs(mid + 1, right, time, totalTrips);
    }

    private boolean isPossible(int[] time, long mid, long totalTrips) {
        long count = 0;
        for (int i : time) {
            count += mid / i;
            if (count >= totalTrips) {
                return true;
            }
        }
        return false;
    }
}