You are given an array time
where time[i]
denotes the time taken by the ith
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.
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.
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.
1 <= time.length <= 105
1 <= time[i], totalTrips <= 107
impl Solution {
pub fn minimum_time(time: Vec<i32>, total_trips: i32) -> i64 {
let total_trips = total_trips as i64;
let mut l = 1_i64;
let mut r = total_trips * *time.iter().min().unwrap() as i64;
while l < r {
let m = (l + r) / 2;
let trips = time.iter().map(|&t| m / t as i64).sum::<i64>();
if trips < total_trips {
l = m + 1;
} else {
r = m;
}
}
r
}
}