Minimize Maximum Distance Between Gas Stations
Binary Search Optimization

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Median of Two Sorted Arrays of Different Sizes

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Problem Examples Solution Visualization Algorithm Code Time Space

Problem Statement

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.

Examples❯

Gas Stations	k	Minimized Max Distance	Description
[1, 2, 3, 4, 5]	1	1.000000	All gaps are equal (1), adding 1 station doesn’t improve max gap
[1, 2, 3, 4, 5]	2	0.666667	We can split two of the 1-length gaps to reduce max gap
[10, 20, 30]	1	10.000000	We split either gap [10,20] or [20,30], max gap becomes 10
[10, 30]	2	6.666667	Gap is 20, splitting into 3 parts gives max 6.66
[0, 100]	9	10.000000	We can add 9 stations to divide 100 into 10 segments
[0, 100]	0	100.000000	No stations added, max distance is the initial gap
[]	2	0.000000	No stations exist, no meaningful distance to minimize
[5]	3	0.000000	Only one station means no gaps, so max distance is 0

Solution❯

This problem challenges us to minimize the longest distance between any two gas stations after adding k new ones. Let’s explore this in a more intuitive way for beginners.

Understanding the Problem

Imagine you’re driving along a highway and gas stations are placed at certain positions. Some parts of the highway may have long gaps between stations. If you’re allowed to place new gas stations, you would want to put them in the widest gaps to make your journey smoother by reducing the largest distance you’d need to travel without fuel.

Goal

After adding k new gas stations, the goal is to make the largest gap as small as possible. This is not necessarily about equalizing all distances — it's about shrinking the biggest one as much as possible.

Key Observations

The array is sorted, so we can easily compute all existing gaps (differences between consecutive stations).
Each gap can be split into smaller segments by placing stations in between.
To minimize the largest segment after splitting, we should distribute the k stations into the largest gaps.

What Does a Valid Answer Look Like?

If a number d is a valid answer, that means we can place k or fewer stations in such a way that no distance between any two gas stations exceeds d. So we're searching for the smallest possible such d.

Different Scenarios

No stations at all: If the array is empty, there's no gap to minimize. We return 0.000000.
Only one station: Again, no gaps exist, so the max distance is 0.
k = 0: We cannot add any station, so the answer is simply the longest existing gap.
Enough stations: If we have many stations, we can split the longest gaps multiple times. For example, a 100-length gap can be split into 10 parts with 9 stations — making the max distance 10.

Approach to Solve

We apply binary search on the answer (the possible distance). We start with the range of [0, max gap] and repeatedly check whether it is possible to achieve a certain distance by placing ≤ k stations. If possible, we try smaller values; if not, we go higher. We stop when we’ve narrowed the answer to a small precision like 1e-6.

This technique is often called binary search with greedy checking — binary search helps us find the minimum possible value, and greedy logic verifies if our guess is feasible.

The result is a floating point number with six decimal precision representing the minimum possible maximum distance between any two stations.

Visualization

Algorithm Steps

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

Code

Python

Java

JavaScript

C++

Kotlin

Swift

Php

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)  # Number of stations needed in this gap
    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  # Try to minimize max distance further
        else:
            low = mid

    return round(high, 6)

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

Time Complexity

Case	Time Complexity	Explanation
Best Case	O(n × log(max_gap/eps))	We perform binary search over precision range and for each guess, count stations in O(n).
Average Case	O(n × log(max_gap/eps))	Each step of binary search tests feasibility by iterating through the array.
Worst Case	O(n × log(max_gap/eps))	In the worst case, we perform many precision steps, each involving scanning the entire array.

Space Complexity

O(1)

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

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Median of Two Sorted Arrays of Different Sizes

Overview of Techniques14
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Sorting6
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Arrays32
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Arrays - Binary Search28
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Strings18
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Matrix3
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Matrix - Binary Search5
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Binary Trees36
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Binary Search Trees14
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Graphs46
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Tries1
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Minimize Maximum Distance Between Gas Stations
Binary Search Optimization

Problem Statement

Examples❯

Solution❯

Understanding the Problem

Goal

Key Observations

What Does a Valid Answer Look Like?

Different Scenarios

Approach to Solve

Visualization

Algorithm Steps

Code

Time Complexity

Space Complexity

Support ProgramGuru.org❯

Minimize Maximum Distance Between Gas StationsBinary Search Optimization

Problem Statement

Examples❯

Solution❯

Understanding the Problem

Goal

Key Observations

What Does a Valid Answer Look Like?

Different Scenarios

Approach to Solve

Visualization

Algorithm Steps

Code

Time Complexity

Space Complexity

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Minimize Maximum Distance Between Gas Stations
Binary Search Optimization