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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.

 

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.

Companies: PhonePe, Google

Related Topics:
Array, Binary Search

Similar Questions:

Solution 1. Binary Answer

// 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;
    }
};