Minimize Max Distance to Gas Station

Problem Statement

In this problem, you are given an integer array named stations which represents the locations of gas stations on a straight line (denoted as the x-axis). Additionally, you are provided with an integer k, indicating the number of new gas stations you are permitted to introduce along this line. The placement of new gas stations is not confined to integer coordinates; they can be situated at any point on the x-axis.

The goal is to strategically add these k gas stations to minimize the maximum distance between any two consecutive gas stations—this maximum distance after placement is termed as the penalty(). The solution requires determining the smallest possible value of this penalty() after all new stations are added, adhering to a precision within 10^-6 of the exact answer.

Examples

Example 1

Input:

stations = [1,2,3,4,5,6,7,8,9,10], k = 9

Output:

0.50000

Example 2

Input:

stations = [23,24,36,39,46,56,57,65,84,98], k = 1

Output:

14.00000

Constraints

• 10 <= stations.length <= 2000
• 0 <= stations[i] <= 108
• stations is sorted in a strictly increasing order.
• 1 <= k <= 106

Approach and Intuition

1. Understanding the distance between stations: The initial step involves identifying gaps between consecutive stations as defined by the stations array. These gaps are potential candidates for minimizing through the strategic placement of new stations.

2. Binary Search for Optimal Penalty:

   • The approach encompasses a binary search on the potential values of penalty(). The lower bound (low) of the search can start from 0 (theoretically placing stations extremely close), and the upper bound (high) can be initialized to the distance between the first and the last station in the list.
   • For each midpoint (mid) in this binary search, which represents a candidate for the smallest maximum distance (penalty), attempt to simulate the placement of new stations. For each gap:
     • Calculate the number of new stations required to ensure that no subsegment within the gap exceeds the distance mid.
     • If the total number of new stations required across all gaps exceeds k, it indicates that mid is too small, and thus, we adjust our search range accordingly.

3. Iterating to Find Minimization:

   • Through the course of the binary search, if a penalty value (midpoint) necessitates fewer or exactly k new stations to achieve the proposed subsegment distances, it is tagged as feasible.
   • Adjust the binary range to zero in on potentially smaller feasible penalties, continuing until the range converges sufficiently close to the minimum penalty, achieving the desired precision.

Example Analysis:

• For example 1, the equidistant initial setup allows for half-unit placements of 9 new stations to tightly pack and minimize the maximum distance, reflecting in the optimal penalty().
• Example 2 offers fewer new stations relative to gaps, exemplifying how a solitary new station influences the longest gap, which determines the penalty() in scenarios of limited placements.

This binary search within specified precision limits provides an efficient pathway to approach and solve the problem, adequately handling the constraints and conditions laid out.

Solutions

class Solution {
    public double findMinDist(int[] position, int Mk) {
        double min = 0, max = 1e8;
        while (max - min > 1e-6) {
            double mid = (min + max) / 2.0;
            if (canDistribute(mid, position, Mk))
                max = mid;
            else
                min = mid;
        }
        return min;
    }

    public boolean canDistribute(double dist, int[] position, int Mk) {
        int count = 0;
        for (int i = 0; i < position.length - 1; ++i)
            count += (int) ((position[i+1] - position[i]) / dist);
        return count <= Mk;
    }
}