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 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.

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.

Approach

Links

• Binarysearch

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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

Example 3

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int minSpeedOnTime(int[] dist, double hour) {
        int n = dist.length, result = -1;
        if(n > 0) {
            int low = 1, high = 10000000;

            while(low <= high) {
                int mid = low + (high-low)/2;
                if(canReach(dist, hour, mid)) {
                    high = mid - 1;
                    result = mid;
                } else {
                    low = mid + 1;
                }
            }
        }
        
        return result;
    }
    
    private boolean canReach(int[] dist, double hour, int speed) {
        double time = 0.0;
        int n = dist.length;
        
        for(int i = 0; i < n-1; i++) {
            time += (int) Math.ceil(dist[i] * 1.0 / speed);
        }
        
        time += (dist[n-1] * 1.0 / speed);
        
        return time <= hour;
    }
}

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