Problem
You are given an array of points in the X-Y plane points where points[i] = [xi, yi].
Return **the minimum area of any rectangle formed from these points, with sides *not necessarily parallel* to the X and Y axes**. If there is not any such rectangle, return 0.
Answers within 10-5 of the actual answer will be accepted.
Example 1:
Input: points = [[1,2],[2,1],[1,0],[0,1]]
Output: 2.00000
Explanation: The minimum area rectangle occurs at [1,2],[2,1],[1,0],[0,1], with an area of 2.
Example 2:
Input: points = [[0,1],[2,1],[1,1],[1,0],[2,0]]
Output: 1.00000
Explanation: The minimum area rectangle occurs at [1,0],[1,1],[2,1],[2,0], with an area of 1.
Example 3:
Input: points = [[0,3],[1,2],[3,1],[1,3],[2,1]]
Output: 0
Explanation: There is no possible rectangle to form from these points.
Constraints:
1 <= points.length <= 50points[i].length == 20 <= xi, yi <= 4 * 10^4All the given points are unique.
Solution (Java)
class Solution {
public double minAreaFreeRect(int[][] points) {
Map<Integer, Set<Integer>> map = new HashMap<>();
double area;
for (int[] point : points) {
map.putIfAbsent(point[0], new HashSet<>());
map.get(point[0]).add(point[1]);
}
double minArea = Double.MAX_VALUE;
int n = points.length;
for (int i = 0; i < n - 2; i++) {
for (int j = i + 1; j < n - 1; j++) {
int dx1 = points[j][0] - points[i][0];
int dy1 = points[j][1] - points[i][1];
// get the 3rd point
for (int k = j + 1; k < n; k++) {
int dx2 = points[k][0] - points[i][0];
int dy2 = points[k][1] - points[i][1];
if (dx1 * dx2 + dy1 * dy2 != 0) {
continue;
}
// find the 4th point
int x = dx1 + points[k][0];
int y = dy1 + points[k][1];
area = calculateArea(points, i, j, k);
if (area >= minArea) {
continue;
}
// 4th point exists
if (map.get(x) != null && map.get(x).contains(y)) {
minArea = area;
}
}
}
}
return minArea == Double.MAX_VALUE ? 0.0 : minArea;
}
private double calculateArea(int[][] points, int i, int j, int k) {
int[] first = points[i];
int[] second = points[j];
int[] third = points[k];
return Math.abs(
first[0] * (second[1] - third[1])
+ second[0] * (third[1] - first[1])
+ third[0] * (first[1] - second[1]));
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).