1870. 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.

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.

Solution

If we find the speed k, then we can guarantee that for all speed2 , if speed2 >= k, then the spendTime with speed2 must be smaller than the given hour.

It likely maintains monotonicity in this observation. So we can try to use binary search.

    public int minSpeedOnTime(int[] dist, double hour) {
        int n = dist.length;
        if(hour - n + 1 <= 0) return -1;
        
        int lo = 1;
        int hi = 10000000;
        
        while(lo < hi){

            int mid = lo + (hi-lo)/2;
            if(isStatisfied(dist, hour, mid)){
                hi = mid;
            }else{
                lo = mid + 1;
            }
        }
        return lo;
    }
    
    public boolean isStatisfied(int[] dist, double hour, int speed){
        double spent = 0;
        for(int i = 0; i < dist.length-1; i++) {
            spent += (int)Math.ceil(((double)(dist[i]))/speed);
        } 
        spent += ((double)(dist[dist.length-1]))/speed;

        return spent <= hour;
    }