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:

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> 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());
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