Minimum Speed to Arrive on Time

Problem Statement

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.

Intuition

"Binary search on answer space - Pattern

Find which speed is minimum and if minimum found, eliminate greater range 
If not found check for greater range

"

Basically 

1 2 3 4 5 ..... 1e5
l                h
Check if mid satisfies
if yes then maybe, 1 to mid-1 may be answer, check for that range

else if mid does not satisfies
Check for mid+1 to 1e5

Links

https://leetcode.com/problems/minimum-speed-to-arrive-on-time/description/

Video Links

https://www.youtube.com/watch?v=mV6aB5RPLmo&t=3s&ab_channel=C0deSutra

Approach 1:

Binary Search over entire Answer Space    

C++

class Solution {
public:
    bool ispossible(vector<int>& dist,int speed,double hour){
        double ans=0;
        for(int i=0;i<dist.size();i++){

            double d=dist[i]*1.0/speed;
            // Converting to double for double division

            if(i!=dist.size()-1)
                ans=ans+ceil(d);

            else
                ans+=d;

            if(ans>hour)
                return false;
        }

        if(ans<=hour)
            return true;

       return false;
    }
    int minSpeedOnTime(vector<int>& dist, double hour) {

        int i=1;
        int j=1e7;
        int minspeed=-1;

        while(i<=j){
            int mid=i+(j-i)/2;
            if(ispossible(dist,mid,hour)){
               minspeed=mid;
               j=mid-1;
            }
            else
                i=mid+1;
        }

        return minspeed;
    }
};

Approach 2:

C++

Approach 3:

C++

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