Leetcode 1637. Widest Vertical Area Between Two Points Containing No Points
Given n points on a 2D plane, where points[i] = [xi, yi], return the widest vertical area between two points such that no points are inside the area. A vertical area is defined as an area of width w where no points lie inside the area. Formally, a vertical area is defined by two parallel lines x = a and x = b where there are no points with x-coordinate in the range (a, b), and a < b.
Before diving into the solution, let’s clarify the problem requirements:
xi and yi?
-10^9 <= xi, yi <= 10^9.[xi, yi] is unique as implied by the problem statement.0.To find the widest vertical area, we can leverage the following strategy:
import java.util.Arrays;
public class Solution {
public int maxWidthOfVerticalArea(int[][] points) {
// Edge case: when the number of points is less than 2
if (points.length < 2) {
return 0;
}
// Extracting the x-coordinates
int[] xCoords = new int[points.length];
for (int i = 0; i < points.length; i++) {
xCoords[i] = points[i][0];
}
// Sorting the x-coordinates
Arrays.sort(xCoords);
// Finding the maximum gap between consecutive x-coordinates
int maxGap = 0;
for (int i = 1; i < xCoords.length; i++) {
int gap = xCoords[i] - xCoords[i - 1];
if (gap > maxGap) {
maxGap = gap;
}
}
return maxGap;
}
public static void main(String[] args) {
Solution solution = new Solution();
int[][] points = { {8,7}, {9,9}, {7,4}, {9,7} };
System.out.println(solution.maxWidthOfVerticalArea(points)); // Output: 1
}
}
O(n log n) time complexity where n is the number of points.O(n), as we iterate through the sorted coordinates.Therefore, the overall time complexity is dominated by the sorting step and is O(n log n).
This solution effectively handles the problem by focusing on sorting the x-coordinates and then finding the maximum gap between successive coordinates. This approach ensures that we cover all potential vertical areas efficiently.
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