<p>You are given an array of points in the <strong>X-Y</strong> plane <code>points</code> where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>.</p>

<p>Return <em>the minimum area of any rectangle formed from these points, with sides <strong>not necessarily parallel</strong> to the X and Y axes</em>. If there is not any such rectangle, return <code>0</code>.</p>

<p>Answers within <code>10<sup>-5</sup></code> of the actual answer will be accepted.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/21/1a.png" style="width: 398px; height: 400px;" />
<pre>
<strong>Input:</strong> points = [[1,2],[2,1],[1,0],[0,1]]
<strong>Output:</strong> 2.00000
<strong>Explanation:</strong> The minimum area rectangle occurs at [1,2],[2,1],[1,0],[0,1], with an area of 2.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/22/2.png" style="width: 400px; height: 251px;" />
<pre>
<strong>Input:</strong> points = [[0,1],[2,1],[1,1],[1,0],[2,0]]
<strong>Output:</strong> 1.00000
<strong>Explanation:</strong> The minimum area rectangle occurs at [1,0],[1,1],[2,1],[2,0], with an area of 1.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/22/3.png" style="width: 383px; height: 400px;" />
<pre>
<strong>Input:</strong> points = [[0,3],[1,2],[3,1],[1,3],[2,1]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There is no possible rectangle to form from these points.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= points.length &lt;= 50</code></li>
	<li><code>points[i].length == 2</code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 4 * 10<sup>4</sup></code></li>
	<li>All the given points are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
  double minAreaFreeRect(vector<vector<int>>& points) {
    map<pair<int, int>, vector<pair<int, int>>> m;
    int n = points.size();

    for(int i{0}; i < n; ++i) {
      for(int j{0}; j < n; ++j) {
        if(i != j) {
          int x{points[i][0] - points[j][0]};
          int y{points[i][1] - points[j][1]};
          m[{x, y}].push_back({i, j});
        }
      }
    }

    double result{DBL_MAX};
    for(auto const& item : m) {
      for(int m{0}; m < item.second.size(); ++m) {
        for(int n{m + 1}; n < item.second.size(); ++n) {
          int i{item.second[m].first};
          int j{item.second[m].second};
          int k{item.second[n].first};

          int dx1{points[i][0] - points[j][0]};
          int dy1{points[i][1] - points[j][1]};
          int dx2{points[i][0] - points[k][0]};
          int dy2{points[i][1] - points[k][1]};

          if(dx1 * dx2 + dy1 * dy2 != 0) {
            continue; // 垂直
          }
          double side1 = (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 side2 = (points[i][0] - points[k][0])*(points[i][0] - points[k][0]) + (points[i][1] - points[k][1]) * (points[i][1] - points[k][1]);
          double area{sqrt(side1) * sqrt(side2)};
          result = min(result, area);
        }
      }
    }
    return result == DBL_MAX ? 0 : result;
  }
};
28178