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Formatted question description: https://leetcode.ca/all/774.html

# 774. Minimize Max Distance to Gas Station

Hard

## Description

On a horizontal number line, we have gas stations at positions stations[0], stations[1], ..., stations[N-1], where N = stations.length.

Now, we add K more gas stations so that D, the maximum distance between adjacent gas stations, is minimized.

Return the smallest possible value of D.

Example:

Input: stations = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], K = 9

Output: 0.500000

Note:

1. stations.length will be an integer in range [10, 2000].
2. stations[i] will be an integer in range [0, 10^8].
3. K will be an integer in range [1, 10^6].
4. Answers within 10^-6 of the true value will be accepted as correct.

## Solution

Use binary search. Sort stations and initialize low to 0 and high to the difference between the maximum element in stations and the minimum element in stations. While high - low > 1e-6, do binary search. Each time set mid to be the average of low and high and use mid as a candidate of D to calculate the minimum number of gas stations to be added. If the minimum number of gas stations to be added is less than or equal to K, then set high = mid. Otherwise, set low = mid. Finally, return low.

• class Solution {
public double minmaxGasDist(int[] stations, int K) {
Arrays.sort(stations);
int length = stations.length;
double low = 0, high = stations[length - 1] - stations[0];
while (high - low > 1e-6) {
double mid = (low + high) / 2;
int minStations = 0;
for (int i = 1; i < length; i++)
minStations += (int) ((stations[i] - stations[i - 1]) / mid);
if (minStations <= K)
high = mid;
else
low = mid;
}
return low;
}
}

• // OJ: https://leetcode.com/problems/minimize-max-distance-to-gas-station/
// Time: O(N * (logN + logM))
// Space: O(1)
class Solution {
bool valid(vector<int> &A, double M, int k) {
for (int i = 1; i < A.size(); ++i) {
k -= (int)((A[i] - A[i - 1]) / M);
if (k < 0) return false;
}
return true;
}
public:
double minmaxGasDist(vector<int>& A, int k) {
double L = 0, R = 1e8, eps = 1e-6;
while (R - L >= eps) {
double M = (L + R) / 2;
if (valid(A, M, k)) R = M;
else L = M;
}
return L;
}
};

• class Solution:
def minmaxGasDist(self, stations: List[int], k: int) -> float:
def check(x):
return sum(int((b - a) / x) for a, b in pairwise(stations)) <= k

left, right = 0, 1e8
while right - left > 1e-6:
mid = (left + right) / 2
if check(mid):
right = mid
else:
left = mid
return left


• func minmaxGasDist(stations []int, k int) float64 {
check := func(x float64) bool {
s := 0
for i, v := range stations[:len(stations)-1] {
s += int(float64(stations[i+1]-v) / x)
}
return s <= k
}
var left, right float64 = 0, 1e8
for right-left > 1e-6 {
mid := (left + right) / 2.0
if check(mid) {
right = mid
} else {
left = mid
}
}
return left
}