1870. Minimum Speed to Arrive on Time

Problem

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

/**
 * @param {number[]} dist
 * @param {number} hour
 * @return {number}
 */
var minSpeedOnTime = function(dist, hour) {
    return search(dist, hour, 1, Math.pow(10, 7));
};

var search = function(dist, hour, left, right) {
    while (left <= right) {
        const mid = left + Math.floor((right - left) / 2);
        if (getTime(dist, mid) > hour) {
            left = mid + 1;
        } else {
            if (left === right) return left;
            right = mid;
        }
    }
    return -1;
};

var getTime = function(dist, speed) {
    var res = 0;
    for (var i = 0; i < dist.length; i++) {
        var num = dist[i] / speed;
        if (i !== dist.length - 1) num = Math.ceil(num);
        res += num;
    }
    return res;
};

Explain:

Binary search.

Complexity:

• Time complexity : O(n * log(K)).
• Space complexity : O(n).