Is Minimize Max Distance to Gas Station easy or hard?

LeetCode classifies this problem as Hard because it requires recognizing the binary-search-on-answer pattern and handling floating-point precision correctly. The feasibility logic itself is simple, but identifying the monotonic property is the key challenge.

Minimize Max Distance to Gas Station Python/Java solution

Most implementations follow the same pattern: binary search on the distance range and compute how many stations are required for each guess. The code works similarly across Python, Java, C++, and Go because it relies only on array iteration and floating-point binary search.

How to solve Minimize Max Distance to Gas Station in O(n log W)?

Use binary search over the maximum allowed distance between adjacent stations. For each candidate distance d, iterate through the array and calculate required += floor((stations[i+1] - stations[i]) / d). If the total required stations is ≤ k, the distance is feasible and you shrink the search range. Continue until the range converges to the required precision.

What is the best approach for Minimize Max Distance to Gas Station?

Binary search on the answer combined with a greedy feasibility check is the optimal approach. You guess a maximum allowed distance d and compute how many additional stations are required so every gap becomes ≤ d. If the required stations exceed k, the distance is too small. This runs in O(n log W) time where W is the largest gap between stations.

Is Minimize Max Distance to Gas Station asked at Google/Amazon/Meta?

This problem reflects a common interview pattern used by companies like Google, Amazon, and Meta: binary search on the answer with a feasibility check. Variations of minimizing the maximum value in a continuous range appear frequently in system optimization and load balancing interview questions.

What data structure is used in Minimize Max Distance to Gas Station?

The optimal solution mainly uses arrays and binary search. A feasibility check iterates through the array of station positions and calculates required insertions. Some alternative solutions also use a max heap (priority queue) to repeatedly split the largest gap.

What is the time complexity of Minimize Max Distance to Gas Station?

The optimal binary search solution runs in O(n log W) time and O(1) space. Each binary search step scans all station gaps to compute how many stations are needed for a candidate distance. W represents the maximum distance between two existing stations.

Description Hints 1 Solution Video 21:44

774. Minimize Max Distance to Gas Station

Hard53.9% acceptance2 companiesPremium elsewhere, free here

You are given an integer array stations that represents the positions of the gas stations on the x-axis. You are also given an integer k.

You should add k new gas stations. You can add the stations anywhere on the x-axis, and not necessarily on an integer position.

Let penalty() be the maximum distance between adjacent gas stations after adding the k new stations.

Return the smallest possible value of penalty(). Answers within 10^-6 of the actual answer will be accepted.

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

Asked at 2 companies

Google Amazon

Topics

Array Binary Search

Hints open one at a time

Hint 1

Use a binary search. We'll binary search the monotone function "possible(D) = can we use K or less gas stations to ensure each adjacent distance between gas stations is at most D?"

21:44

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