1870. Minimum Speed to Arrive on Time
Description
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
.
Solutions
Solution: Binary Search
- Time complexity: O(nlogn)
- Space complexity: O(1)
JavaScript
/**
* @param {number[]} dist
* @param {number} hour
* @return {number}
*/
const minSpeedOnTime = function (dist, hour) {
const size = dist.length;
if (size - 1 >= hour) return -1;
const MAX_SPEED = 10 ** 7;
const calculateReachTime = speed => {
let result = 0;
for (let index = 0; index < size - 1; index++) {
result += Math.ceil(dist[index] / speed);
}
return result + dist[size - 1] / speed;
};
let left = 1;
let right = MAX_SPEED;
while (left < right) {
const mid = Math.floor((left + right) / 2);
const reachTime = calculateReachTime(mid);
reachTime > hour ? (left = mid + 1) : (right = mid);
}
return left;
};