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 <= 10^5
1 <= time[i], totalTrips <= 10^7
题解
经过对题目的分析可知,题目要求一个最小阈值,这个最小的阈值能够满足下面的条件:
代码
class Solution {
public:
int n,c;
vector<int>a;
bool isok(long long s){
long long sum=0;
for(int i=0;i<n;i++){
sum+=s/a[i];
}
return sum>=c;
}
long long minimumTime(vector<int>& time, int totalTrips) {
n=time.size();
for(int i=0;i<n;i++){
a.push_back(time[i]);
}
c=totalTrips;
long long MAX=0;
for(int i=0;i<n;i++){
MAX=max(MAX,(long long)time[i]);
}
MAX*=totalTrips;
long long l=0,r=MAX,mid,ans;
while(l<=r){
mid=l+(r-l)/2;
if(isok(mid))
ans=mid,r=mid-1;
else
l=mid+1;
}
return ans;
}
};