Problem Explanation

The problem involves a horizontal number line on which several gas stations exist at certain positions. We are required to add more gas stations (the number to be added is given) such that the maximum distance between any two gas stations is minimized. We have to return this minimum possible distance.

For instance, let's say we have the following gas stations at positions [1,2,3,4,5,6,7,8,9,10] and we can add 9 more stations. In this case, the optimal strategy would be to fill each gap with a new station, thus making the maximum distance 0.5.

Solution Approach

The best approach for solving this problem is to use the binary search method. We want to find the minimum maximum distance between stations after adding k gas stations.

We start by initializing our maximum possible distance, which is equal to the last station minus the first. In the binary search, we create a mid-point and check if the number of gas stations we need to add is less than or equal to k for the given distance. If it is, we decrease the distance, else we increase the distance.

It's important to determine how we can check if a given distance is possible. To find this, we iterate over the stations and calculate the difference between the current and previous station. If this difference exceeds our given distance, we increase the count by ceil(diff / mid) - 1.

If the count is less than k, it means that it is possible to achieve a smaller distance. Hence we update our r variable to mid. If the count is more than k, it means we need more stations than available to achieve this distance. Hence we update our l variable to mid. We continue this process till the difference between l and r is more than a very small number close to 0 (1e-6 in this case).

Walkthrough of an example:

Consider stations = [1,2,3,4,5,9] and k=4. Let's see how the binary search method will work here.

• Initially, l=0 and r = 9-1=8.
• In the first iteration, mid=4. Count the stations necessary to ensure the maximum difference does not exceed 4. Between 1 and 2, 2 and 3, 3 and 4, no station is needed. Between 4 and 5, one station is needed. Between 5 and 9, one station is needed. So total count=2 which is less than 4. Hence we should reduce our distance. So r=mid=4.
• Now l=0 and r=4. Mid=2. Count the stations necessary at this distance. Between each pair of existing stations, one station will be needed. So total count=5 which is more than 4. Hence our distance needs to increase. So l=mid=2.
• Now l=2 and r=4. Mid=3. Count the stations necessary at this distance. Between 1 and 2, 2 and 3, 3 and 4, no station is needed. Between 4 and 5, one station is needed. Between 5 and 9, one is needed. So total count=2 which is less than 4. Hence we should reduce our distance. So r=mid=3.
• Now l=2 and r=3. Mid=2.5. Count the stations necessary at this distance. Between each pair of existing stations, one station will be needed. So total count=5 which is more than 4. Hence our distance needs to increase. So l=mid=2.5.
• Now l=2.5 and r=3. Mid=2.75. Count the stations necessary at this distance. Between 1 and 2, 2 and 3, 3 and 4, no station is needed. Between 4 and 5, one station is needed. Between 5 and 9, 2 will be needed. So total count=3 which is less than 4. Hence we should reduce our distance. So r=mid=2.75.
• Now l=2.5 and r=2.75. Mid=2.625. Count the stations necessary at this distance. Between 1 and 2, 2 and 3, 3 and 4, no station is needed. Between 4 and 5, one station is needed. Between 5 and 9, 2 will be needed. So total count=3 which is less than 4. Hence we should reduce our distance. So r=mid=2.625.
• We continue this process till the difference between l and r is more than a small number close to 0.

Solution in Python

1
2python
3class Solution:
4    def minmaxGasDist(self, stations: List[int], K: int) -> float:
5        def possible(D):
6            return sum(int((stations[i+1]-stations[i]) / D) for i in range(len(stations)-1)) <= K
7
8        l, r = 0, 1e8
9
10        while l < r - 1e-6:
11            m = (l + r) / 2
12            if possible(m):
13                r = m
14            else:
15                l = m    
16
17        return r

Solution in Java

1
2java
3class Solution {
4    public double minmaxGasDist(int[] stations, int K) {
5		double l = 0, r = 1e8;
6		
7		while (r - l > 1e-6) {
8			double mid = (l + r) / 2.0;
9			int count = 0;
10			for (int i = 0; i < stations.length - 1; i++)
11				count += Math.ceil((stations[i + 1] - stations[i]) / mid) - 1;
12			if (count > K) l = mid;
13			else r = mid;
14		}
15		return r;
16	}
17}

Solution in JavaScript

1
2javascript
3var minmaxGasDist = function(stations, K) {
4    let lo = 0, hi = 1e8;
5    while (hi-lo > 1e-6) {
6        let mi = (lo + hi) / 2.0;
7        let need = 0;
8        for (let i = 0; i < stations.length-1; ++i)
9            need += Math.ceil((stations[i+1] - stations[i]) / mi) - 1;
10        if (need > K)
11            lo = mi;
12        else
13            hi = mi;
14    }
15    return lo;
16};

Solution in C++

1
2cpp
3class Solution {
4public:
5    double minmaxGasDist(vector<int>& stations, int K) {
6        double l = 0, r = 1e8;
7        while (r - l > 1e-6) {
8            double m = (l + r) / 2.0;
9            int need = 0;
10            for (int i = 0; i < stations.size()-1; ++i)
11                need += ceil((stations[i+1] - stations[i]) / m) - 1;
12            if (need > K)
13                l = m;
14            else
15                r = m;
16        }
17        return l;
18    }
19};

Solution in C#

1
2csharp
3public class Solution {
4    public double MinmaxGasDist(int[] stations, int K) {
5        double l = 0, h = 1e8;
6        while (h - l > 1e-6) {
7            double m = 0.5 * (l + h);
8            int need = 0;
9            for (int i = 0; i < stations.Length - 1; ++i)
10                need += (int)((stations[i+1] - stations[i]) / m);
11            if (need > K) l = m;
12            else h = m;
13        }
14        return l;
15    }
16}

Space Complexity

In terms of space complexity, all solutions are quite efficient as they only need enough storage for the gas stations and a few extra variables. Hence, the space complexity is O(n), where n is the number of gas stations.

Time Complexity

The time complexity of all the solutions is O(n log m), where n is the number of gas stations and m is the range within which the gas stations are distributed. This complexity arises because in the worst-case scenario, we iterate over the entire list of stations for each step in our binary search.

Optimizations

There is not much scope for optimization in this particular problem as the solution already uses an efficient binary search algorithm. However, an improvement in code efficiency could be achieved by reducing the precision required for the minimum difference between low and high to a larger value (greater than 1e-6). This would reduce the number of binary search iterations but at the cost of a less precise result.

Another possible optimization is using integer division instead of floating division for calculating the number of extra stations needed in each interval. Then instead of using the Math.ceil() function for rounding up, we would just add 1 if there is a remainder. However, this would again reduce the precision of the result and might not be applicable for all test cases.

In order to ensure precise results, it is recommended to stick to the provided solution.