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
1sttrain ride takes1.5hours, you must wait for an additional0.5hours before you can depart on the2ndtrain 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.length1 <= n <= 1051 <= dist[i] <= 1051 <= hour <= 109- There will be at most two digits after the decimal point in
hour.
Solution: Binary Search
l = speedmin=1
r = speedmax+1 = 1e7 + 1
Find the min valid speed m such that t(m) <= hour.
Time complexity: O(nlogn)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: int minSpeedOnTime(vector<int>& dist, double hour) { constexpr int kMax = 1e7 + 1; const int n = dist.size(); int l = 1; int r = kMax; while (l < r) { int m = l + (r - l) / 2; int t = 0; for (int i = 0; i < n - 1; ++i) t += (dist[i] + m - 1) / m; if (t + dist.back() * 1.0 / m <= hour) r = m; else l = m + 1; } return l == kMax ? -1 : l; } }; |