1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
|
// 1870. Minimum Speed to Arrive on Time
#include "leetcode.h"
/*
* 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 kilometres of the i'th train ride.
* each train can only depart at an integer hour, so you may need to wait in
* between each ride. return the minimum number speed in km/h that all the
* trains must travel at for you to reach the office on time, or '-1' if it is
* possible to be on time.
*/
class Solution {
bool can_reach_on_time(vector<int> &dist, double hour, int speed) {
double time = 0;
for (int i = 0; i < dist.size() - 1; ++i)
time += ((dist[i] + speed - 1) / speed);
time += ((double)dist.back()) / speed;
return time <= hour;
}
public:
int minSpeedOnTime(vector<int> dist, double hour) {
int n = dist.size();
if (hour <= n - 1)
return -1;
int low = 1, high = 1e7, mid;
while (low < high) {
mid = (low + high) / 2;
if (can_reach_on_time(dist, hour, mid))
high = mid;
else
low = mid + 1;
}
return high;
}
};
int main() {
Solution obj;
vector<int> dist = {1, 3, 2};
printf("%d\n", obj.minSpeedOnTime(dist, 6)); // expect: 1
printf("%d\n", obj.minSpeedOnTime(dist, 2.7)); // expect: 3
printf("%d\n", obj.minSpeedOnTime(dist, 1.9)); // expect: -1
}
|