1870. Minimum Speed to Arrive on Time LeetCode Solution
In this guide, you will get 1870. Minimum Speed to Arrive on Time LeetCode Solution with the best time and space complexity. The solution to Minimum Speed to Arrive on Time problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Minimum Speed to Arrive on Time
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 takes 1.5 hours, you must wait for an additional 0.5 hours before you can depart on the 2nd 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.
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.
Complexity Analysis
Time Complexity: O(n\log(r – l)), where r = 10^7 and l = 1
Space Complexity: O(1)
1870. Minimum Speed to Arrive on Time LeetCode Solution in C++
class Solution {
public:
int minSpeedOnTime(vector<int>& dist, double hour) {
int ans = -1;
int l = 1;
int r = 1e7;
while (l <= r) {
const int m = (l + r) / 2;
if (time(dist, hour, m) > hour) {
l = m + 1;
} else {
ans = m;
r = m - 1;
}
}
return ans;
}
private:
double time(const vector<int>& dist, double hour, int speed) {
double sum = 0;
for (int i = 0; i < dist.size() - 1; ++i)
sum += ceil((double)dist[i] / speed);
return sum + (double)dist.back() / speed;
}
}; /* code provided by PROGIEZ */
1870. Minimum Speed to Arrive on Time LeetCode Solution in Java
class Solution {
public int minSpeedOnTime(int[] dist, double hour) {
int ans = -1;
int l = 1;
int r = (int) 1e7;
while (l <= r) {
final int m = (l + r) / 2;
if (time(dist, hour, m) > hour) {
l = m + 1;
} else {
ans = m;
r = m - 1;
}
}
return ans;
}
private double time(int[] dist, double hour, int speed) {
double sum = 0;
for (int i = 0; i < dist.length - 1; ++i)
sum += Math.ceil((double) dist[i] / speed);
return sum + (double) dist[dist.length - 1] / speed;
}
} // code provided by PROGIEZ
1870. Minimum Speed to Arrive on Time LeetCode Solution in Python
class Solution:
def minSpeedOnTime(self, dist: list[int], hour: float) -> int:
ans = -1
l = 1
r = int(1e7)
def time(speed: int) -> float:
summ = 0
for i in range(len(dist) - 1):
summ += math.ceil(dist[i] / speed)
return summ + dist[-1] / speed
while l <= r:
m = (l + r) // 2
if time(m) > hour:
l = m + 1
else:
ans = m
r = m - 1
return ans # code by PROGIEZ
