## DEV Community

Abhishek Chaudhary

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# Minimum Time to Complete Trips

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.

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`

SOLUTION:

``````class Solution:
def numTrips(self, time, t):
trips = 0
for el in time:
trips += t // el
return trips

def minimumTime(self, time: List[int], totalTrips: int) -> int:
beg = 1
end = min(time) * totalTrips
while beg <= end:
mid = (beg + end) // 2
currtrips = self.numTrips(time, mid)
nexttrips = self.numTrips(time, mid + 1)
if currtrips < totalTrips and nexttrips >= totalTrips:
return mid + 1
elif beg == end:
return mid
elif currtrips < totalTrips:
beg = mid
else:
end = mid
``````