Minimize Maximum Distance Between Gas Stations

You are given a sorted array of integers representing the positions of existing gas stations along a road. You are also given an integer k, which represents the number of additional gas stations you are allowed to add anywhere between the existing ones.

Your task is to minimize the maximum distance between any two adjacent gas stations after placing the k new stations optimally.

Return the smallest possible maximum distance between gas stations after the optimal placement, rounded to 6 decimal places if required.

Understanding the Problem

You're given an array representing the positions of gas stations along a highway. The goal is to add k additional gas stations in such a way that the maximum distance between any two adjacent gas stations becomes as small as possible.

Why is this important? Imagine you're driving, and there's a really long stretch without a gas station. That’s uncomfortable and risky. So, if you could add a few more stations, where would you place them? Intuitively, you'd want to place them in the longest stretches — that’s what this problem is all about.

Step-by-Step Solution with Example

Step 1: Observe the distances

Let’s take an example: stations = [1, 5, 10] and k = 1. This means there are gas stations at positions 1, 5, and 10, and you're allowed to add 1 more.

The current gaps are:

• Between 1 and 5: distance 4
• Between 5 and 10: distance 5

So, the largest gap is 5. We want to insert a station such that the largest gap after insertion is minimized.

Step 2: Use Binary Search on the answer

Instead of trying all possible placements (which is slow), we can use binary search. The idea is: if we guess a maximum allowed distance d, can we split all the existing gaps using ≤ k stations so that no segment exceeds d?

Step 3: Check if a guess is valid (Greedy check)

For each gap between two stations, say of length len, we can divide it into ceil(len / d) parts. That means we’d need ceil(len / d) - 1 extra stations in that gap.

We repeat this for every gap and count the total stations needed. If it's ≤ k, then our guess d is valid.

Step 4: Perform the binary search

We search for the smallest possible d using binary search. Start with low = 0 and high = max_gap. While high - low > 1e-6, compute mid and check if it’s possible. Adjust the search bounds accordingly.

Step 5: Return the answer

After the binary search ends, low (or high) will hold the minimum possible maximum distance between any two stations after placing k new ones.

Edge Cases

• No stations: If the input array is empty, there’s nothing to minimize. Return 0.000000.
• Only one station: No gaps exist, so return 0.000000.
• k = 0: You can't add any new station. So, the maximum distance is just the biggest gap between existing stations.
• Large k: If you have many stations to add, you can split the longest gaps into very small chunks, bringing the max distance down significantly.

Finally

This problem is a classic case of using binary search on the answer — a clever technique when the answer is a real number instead of an integer. By combining binary search with greedy checking, we efficiently minimize the largest distance between stations. It's not about equalizing all distances but about shrinking the worst one.

Always remember to handle edge cases before diving into implementation, and use precision comparisons carefully when working with floating-point numbers.

1. Initialize low = 0 and high = max(arr[i+1] - arr[i]).
2. While high - low > 1e-6:
3. → Compute mid = (low + high) / 2.
4. → Check if it's possible to place ≤ k stations so that max distance ≤ mid.
5. → If yes, update high = mid.
6. → Else, update low = mid.
7. Return high (smallest valid max distance).

#include <stdio.h>
#include <math.h>

int isPossible(int arr[], int n, int k, double dist) {
    int count = 0;
    for (int i = 0; i < n - 1; i++) {
        double gap = arr[i+1] - arr[i];
        count += (int)(gap / dist);
    }
    return count <= k;
}

double minimizeMaxDistance(int arr[], int n, int k) {
    double low = 0.0, high = 0.0, eps = 1e-6;
    for (int i = 0; i < n - 1; i++) {
        double gap = arr[i+1] - arr[i];
        if (gap > high) high = gap;
    }
    while (high - low > eps) {
        double mid = (low + high) / 2;
        if (isPossible(arr, n, k, mid)) {
            high = mid;
        } else {
            low = mid;
        }
    }
    return round(high * 1000000) / 1000000;
}

int main() {
    int arr[] = {1, 2, 8};
    int k = 1;
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("Minimum Max Distance: %.6f\n", minimizeMaxDistance(arr, n, k));
    return 0;
}

#include <iostream>
#include <cmath>
using namespace std;

bool isPossible(int arr[], int n, int k, double dist) {
    int count = 0;
    for (int i = 0; i < n - 1; i++) {
        double gap = arr[i+1] - arr[i];
        count += (int)(gap / dist);
    }
    return count <= k;
}

double minimizeMaxDistance(int arr[], int n, int k) {
    double low = 0.0, high = 0.0, eps = 1e-6;
    for (int i = 0; i < n - 1; i++) {
        high = max(high, (double)(arr[i+1] - arr[i]));
    }
    while (high - low > eps) {
        double mid = (low + high) / 2;
        if (isPossible(arr, n, k, mid)) {
            high = mid;
        } else {
            low = mid;
        }
    }
    return round(high * 1e6) / 1e6;
}

int main() {
    int arr[] = {1, 2, 8};
    int k = 1;
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Minimum Max Distance: " << minimizeMaxDistance(arr, n, k) << endl;
    return 0;
}

def is_possible(arr, k, dist):
    count = 0
    for i in range(len(arr) - 1):
        gap = arr[i+1] - arr[i]
        count += int(gap // dist)
    return count <= k

def minimize_max_distance(arr, k):
    low = 0.0
    high = max(arr[i+1] - arr[i] for i in range(len(arr)-1))
    eps = 1e-6

    while high - low > eps:
        mid = (low + high) / 2
        if is_possible(arr, k, mid):
            high = mid
        else:
            low = mid

    return round(high, 6)

arr = [1, 2, 8]
k = 1
print("Minimum Max Distance:", minimize_max_distance(arr, k))

public class MinimizeMaxDistance {
    public static boolean isPossible(int[] arr, int k, double dist) {
        int count = 0;
        for (int i = 0; i < arr.length - 1; i++) {
            double gap = arr[i+1] - arr[i];
            count += (int)(gap / dist);
        }
        return count <= k;
    }

    public static double minimizeMaxDistance(int[] arr, int k) {
        double low = 0, high = 0;
        for (int i = 0; i < arr.length - 1; i++) {
            high = Math.max(high, arr[i+1] - arr[i]);
        }

        double eps = 1e-6;
        while (high - low > eps) {
            double mid = (low + high) / 2;
            if (isPossible(arr, k, mid)) {
                high = mid;
            } else {
                low = mid;
            }
        }
        return Math.round(high * 1e6) / 1e6;
    }

    public static void main(String[] args) {
        int[] arr = {1, 2, 8};
        int k = 1;
        System.out.println("Minimum Max Distance: " + minimizeMaxDistance(arr, k));
    }
}

function isPossible(arr, k, dist) {
  let count = 0;
  for (let i = 0; i < arr.length - 1; i++) {
    let gap = arr[i + 1] - arr[i];
    count += Math.floor(gap / dist);
  }
  return count <= k;
}

function minimizeMaxDistance(arr, k) {
  let low = 0;
  let high = Math.max(...arr.map((_, i) => i < arr.length - 1 ? arr[i + 1] - arr[i] : 0));
  let eps = 1e-6;

  while (high - low > eps) {
    let mid = (low + high) / 2;
    if (isPossible(arr, k, mid)) {
      high = mid;
    } else {
      low = mid;
    }
  }
  return Number(high.toFixed(6));
}

console.log("Minimum Max Distance:", minimizeMaxDistance([1, 2, 8], 1));

package main

import (
    "fmt"
    "math"
)

func isPossible(arr []int, k int, dist float64) bool {
    count := 0
    for i := 0; i < len(arr)-1; i++ {
        gap := float64(arr[i+1] - arr[i])
        count += int(gap / dist)
    }
    return count <= k
}

func minimizeMaxDistance(arr []int, k int) float64 {
    low := 0.0
    high := 0.0
    eps := 1e-6
    for i := 0; i < len(arr)-1; i++ {
        gap := float64(arr[i+1] - arr[i])
        if gap > high {
            high = gap
        }
    }
    for high - low > eps {
        mid := (low + high) / 2
        if isPossible(arr, k, mid) {
            high = mid
        } else {
            low = mid
        }
    }
    return math.Round(high*1e6) / 1e6
}

func main() {
    arr := []int{1, 2, 8}
    k := 1
    fmt.Printf("Minimum Max Distance: %.6f\n", minimizeMaxDistance(arr, k))
}

fn is_possible(arr: &Vec<i32>, k: i32, dist: f64) -> bool {
    let mut count = 0;
    for i in 0..arr.len()-1 {
        let gap = (arr[i+1] - arr[i]) as f64;
        count += (gap / dist).floor() as i32;
    }
    count <= k
}

fn minimize_max_distance(arr: &Vec<i32>, k: i32) -> f64 {
    let mut low = 0.0;
    let mut high = (1..arr.len()).map(|i| arr[i] - arr[i-1]).map(|x| x as f64).fold(0.0, f64::max);
    let eps = 1e-6;
    while high - low > eps {
        let mid = (low + high) / 2.0;
        if is_possible(arr, k, mid) {
            high = mid;
        } else {
            low = mid;
        }
    }
    (high * 1e6).round() / 1e6
}

fn main() {
    let arr = vec![1, 2, 8];
    let k = 1;
    println!("Minimum Max Distance: {:.6}", minimize_max_distance(&arr, k));
}

fun isPossible(arr: IntArray, k: Int, dist: Double): Boolean {
    var count = 0
    for (i in 0 until arr.size - 1) {
        val gap = arr[i + 1] - arr[i]
        count += (gap / dist).toInt()
    }
    return count <= k
}

fun minimizeMaxDistance(arr: IntArray, k: Int): Double {
    var low = 0.0
    var high = arr.zipWithNext { a, b -> (b - a).toDouble() }.maxOrNull() ?: 0.0
    val eps = 1e-6

    while (high - low > eps) {
        val mid = (low + high) / 2
        if (isPossible(arr, k, mid)) {
            high = mid
        } else {
            low = mid
        }
    }
    return String.format("%.6f", high).toDouble()
}

fun main() {
    val arr = intArrayOf(1, 2, 8)
    val k = 1
    println("Minimum Max Distance: ${minimizeMaxDistance(arr, k)}")
}

import Foundation

func isPossible(_ arr: [Int], _ k: Int, _ dist: Double) -> Bool {
    var count = 0
    for i in 0..<arr.count - 1 {
        let gap = Double(arr[i+1] - arr[i])
        count += Int(gap / dist)
    }
    return count <= k
}

func minimizeMaxDistance(_ arr: [Int], _ k: Int) -> Double {
    var low = 0.0
    var high = Double(arr.enumerated().dropFirst().map { $1 - arr[$0.offset - 1] }.max() ?? 0)
    let eps = 1e-6

    while high - low > eps {
        let mid = (low + high) / 2
        if isPossible(arr, k, mid) {
            high = mid
        } else {
            low = mid
        }
    }
    return (high * 1e6).rounded() / 1e6
}

let arr = [1, 2, 8]
let k = 1
print("Minimum Max Distance: \(minimizeMaxDistance(arr, k))")

function isPossible(arr: number[], k: number, dist: number): boolean {
  let count = 0;
  for (let i = 0; i < arr.length - 1; i++) {
    let gap = arr[i + 1] - arr[i];
    count += Math.floor(gap / dist);
  }
  return count <= k;
}

function minimizeMaxDistance(arr: number[], k: number): number {
  let low = 0;
  let high = Math.max(...arr.map((_, i) => i < arr.length - 1 ? arr[i + 1] - arr[i] : 0));
  let eps = 1e-6;

  while (high - low > eps) {
    let mid = (low + high) / 2;
    if (isPossible(arr, k, mid)) {
      high = mid;
    } else {
      low = mid;
    }
  }
  return Number(high.toFixed(6));
}

console.log("Minimum Max Distance:", minimizeMaxDistance([1, 2, 8], 1));

Time complexity

Space complexity

O(1)

No extra space is used apart from a few variables for binary search and counters.