You are given a floating-point number hour
, representing the amount of time you have to reach the office. To commute to the office, you must take n
trains in sequential order. You are also given an integer array dist
of length n
, where dist[i]
describes the distance (in kilometers) of the ith
train ride.
Each train can only depart at an integer hour, so you may need to wait in between each train ride.
- For example, if the
1st
train ride takes1.5
hours, you must wait for an additional0.5
hours before you can depart on the2nd
train ride at the 2 hour mark.
Return the minimum positive integer speed (in kilometers per hour) that all the trains must travel at for you to reach the office on time, or -1
if it is impossible to be on time.
Tests are generated such that the answer will not exceed 107
and hour
will have at most two digits after the decimal point.
Example 1:
Input: dist = [1,3,2], hour = 6 Output: 1 Explanation: At speed 1: - The first train ride takes 1/1 = 1 hour. - Since we are already at an integer hour, we depart immediately at the 1 hour mark. The second train takes 3/1 = 3 hours. - Since we are already at an integer hour, we depart immediately at the 4 hour mark. The third train takes 2/1 = 2 hours. - You will arrive at exactly the 6 hour mark.
Example 2:
Input: dist = [1,3,2], hour = 2.7 Output: 3 Explanation: At speed 3: - The first train ride takes 1/3 = 0.33333 hours. - Since we are not at an integer hour, we wait until the 1 hour mark to depart. The second train ride takes 3/3 = 1 hour. - Since we are already at an integer hour, we depart immediately at the 2 hour mark. The third train takes 2/3 = 0.66667 hours. - You will arrive at the 2.66667 hour mark.
Example 3:
Input: dist = [1,3,2], hour = 1.9 Output: -1 Explanation: It is impossible because the earliest the third train can depart is at the 2 hour mark.
Constraints:
n == dist.length
1 <= n <= 105
1 <= dist[i] <= 105
1 <= hour <= 109
- There will be at most two digits after the decimal point in
hour
.
Related Topics:
Array, Binary Search
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// OJ: https://leetcode.com/problems/minimum-speed-to-arrive-on-time/
// Author: github.com/lzl124631x
// Time: O(NlogT) where T is the maximum possible speed
// Space: O(1)
class Solution {
public:
int minSpeedOnTime(vector<int>& A, double hour) {
long L = 1, R = 1e7, N = A.size();
if (hour < N - 1) return -1;
auto valid = [&](int speed) {
double time = 0;
for (int n : A) {
time = ceil(time);
time += (double)n / speed;
if (time > hour) return false;
}
return true;
};
while (L <= R) {
long M = (L + R) / 2;
if (valid(M)) R = M - 1;
else L = M + 1;
}
return L > 1e7 ? -1 : L;
}
};