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This approach involves sorting the x-coordinates of the points and then finding the maximum difference between consecutive sorted x-coordinates. This difference gives the widest vertical area between any two points.
Time Complexity: O(n log n), due to the sorting step.
Space Complexity: O(n), due to the additional array holding the sorted x-coordinates.
1import java.util.Arrays;
2
3public class Solution {
4 public int maxWidthOfVerticalArea(int[][] points) {
5 int[] xCoords = new int[points.length];
6 for (int i = 0; i < points.length; i++) {
7 xCoords[i] = points[i][0];
8 }
9 Arrays.sort(xCoords);
10 int maxWidth = 0;
11 for (int i = 1; i < xCoords.length; i++) {
12 maxWidth = Math.max(maxWidth, xCoords[i] - xCoords[i - 1]);
13 }
14 return maxWidth;
15 }
16
17 public static void main(String[] args) {
18 Solution solution = new Solution();
19 int[][] points = {{3, 1}, {9, 0}, {1, 0}, {1, 4}, {5, 3}, {8, 8}};
20 System.out.println(solution.maxWidthOfVerticalArea(points));
21 }
22}
In this Java implementation, a separate array is used to extract x-coordinates. The Arrays.sort()
method sorts this array, and a loop computes the widest vertical area.
Although not commonly necessary due to the constraints provided, if x-coordinates were to be over a smaller range, a variant of bucket sort could be used to place x-coordinates in a collection of buckets. This helps find consecutive x-coordinate gaps in nearly constant time, potentially improving performance by eliminating the general-purpose sort with more specialized data placement.
Time Complexity: O(n) expected when bucket conditions are met.
Space Complexity: O(m) where m is the bucket range size.
Explores data distribution rather than generic sorting for highly regular input domains.