2023-03-07 Daily Challenge
Today I have done leetcode's March LeetCoding Challenge with cpp
.
March LeetCoding Challenge 7
Description
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 {
public:
long long minimumTime(vector<int>& time, int totalTrips) {
long long high = 1e15;
long long low = 1;
while(low < high) {
long long mid = (high + low) >> 1;
long long total = 0;
for(auto time : time) {
total += mid / time;
if(total > INT_MAX) {
break;
}
}
if(total < totalTrips) {
low = mid + 1;
} else {
high = mid;
}
}
return low;
}
};
// Accepted
// 123/123 cases passed (270 ms)
// Your runtime beats 86.49 % of cpp submissions
// Your memory usage beats 33.4 % of cpp submissions (94.6 MB)