963. Minimum Area Rectangle II

Given a set of points in the xy-plane, determine the minimum area of any rectangle formed from these points, with sides not necessarily parallel to the x and y axes.

If there isn't any rectangle, return 0.

Example 1:

Input: [[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: [[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: [[0,3],[1,2],[3,1],[1,3],[2,1]]
Output: 0
Explanation: There is no possible rectangle to form from these points.

Example 4:

Input: [[3,1],[1,1],[0,1],[2,1],[3,3],[3,2],[0,2],[2,3]]
Output: 2.00000
Explanation: The minimum area rectangle occurs at [2,1],[2,3],[3,3],[3,1], with an area of 2.

Note:

1. 1 <= points.length <= 50
2. 0 <= points[i][0] <= 40000
3. 0 <= points[i][1] <= 40000
4. All points are distinct.
5. Answers within 10^-5 of the actual value will be accepted as correct.

Approach #1: Math. [Java]

class Solution {
    public double minAreaFreeRect(int[][] points) {
        int len = points.length;
        if (len < 4) return 0.0;
        double ret = Double.MAX_VALUE;
        Map<String, List<int[]>> map = new HashMap<>();
        for (int i = 0; i < len; ++i) {
            for (int j = i+1; j < len; ++j) {
                long diagonal = (points[i][0] - points[j][0]) * (points[i][0] - points[j][0]) +
                               (points[i][1] - points[j][1]) * (points[i][1] - points[j][1]);
                double centerX = (double)(points[i][0] + points[j][0]) / 2;
                double centerY = (double)(points[i][1] + points[j][1]) / 2;
                String key = "" + diagonal + "+" + centerX + "+" + centerY;
                if (map.get(key) == null) map.put(key, new ArrayList<int[]>());
                map.get(key).add(new int[]{i, j});
            }
        }

        for (String key : map.keySet()) {
            List<int[]> list = map.get(key);
            if (list.size() < 2) continue;
            for (int i = 0; i < list.size(); ++i) {
                for (int j = i+1; j < list.size(); ++j) {
                    int p1 = list.get(i)[0];
                    int p2 = list.get(j)[0];
                    int p3 = list.get(j)[1];
                    double x = Math.sqrt((points[p1][0] - points[p2][0]) * (points[p1][0] - points[p2][0])
                                    + (points[p1][1] - points[p2][1]) * (points[p1][1] - points[p2][1]));
                    double y = Math.sqrt((points[p1][0] - points[p3][0]) * (points[p1][0] - points[p3][0])
                                    + (points[p1][1] - points[p3][1]) * (points[p1][1] - points[p3][1]));
                    double area = x * y;
                    ret = Math.min(ret, area);
                }
            }
        }

        return ret == Double.MAX_VALUE ? 0.0 : ret;
    }
}

　　

Analysis:

1. Two diagonals of a rectangle bisect each other, and are of equal length.

2. The map's key is String including diagonal length and coordinate of the diagonal center; map's vlaue is the index of two points forming the diagonal.

Reference:

https://leetcode.com/problems/minimum-area-rectangle-ii/discuss/208361/JAVA-O(n2)-using-Map