712303. Minimize Max Distance to Gas Station
Hard
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Minimize Max Distance to Gas Station
Summary
The problem requires you to find a gas station with the minimum maximum distance from all other gas stations. Given an array of integers representing the positions of gas stations, your goal is to minimize the maximum distance between any two stations.
Tags: Binary Search, Mathematical, Arrays, Data Structures, Algorithms
Difficulty: Hard
Detailed Explanation
The problem involves finding a set of gas stations that are spaced as evenly as possible. To achieve this, we'll use binary search to find the optimal solution.
Here's a step-by-step breakdown of the algorithm:
- First, sort the array of gas station positions in ascending order.
- Initialize two variables:
leftandright. Setleftto 0 (the first element in the sorted array) andrightto the last element in the array. - Perform a binary search to find the optimal distance between gas stations.
- In each iteration of the binary search, calculate the total distance traveled if we were to space the gas stations at a distance equal to
(left + right) / 2. This is calculated by iterating through the sorted array and adding up the distances between consecutive gas stations, using the midpoint as the spacing for all gas stations. - If the total distance traveled in step 4 is less than or equal to
right - left, updaterightto(left + right) / 2. This means that we've found a better spacing, and we should narrow our search range to the left half of the current interval. - Otherwise, update
leftto(left + right) / 2 + 1. This means that we haven't found a better spacing yet, so we should widen our search range to the right half of the current interval. - Repeat steps 4-6 until
right - leftis less than or equal to 1. - The optimal distance between gas stations is now
(left + right) / 2.
Time complexity: O(n log n), where n is the number of gas stations. Space complexity: O(1).
Optimized Solutions
Java
public int minMaxDistance(int[] positions) {
Arrays.sort(positions);
int left = 0, right = positions.length - 1;
while (right - left > 1) {
int mid = (left + right) / 2;
long totalDistance = 0;
for (int i = 0; i < positions.length - 1; i++) {
totalDistance += Math.abs(positions[i + 1] - positions[i]) - mid;
}
if (totalDistance <= 0) {
right = mid;
} else {
left = mid + 1;
}
}
return (int)((left + right) / 2);
}