<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) {
        int n = points.size();
        double ans = (double)INT_MAX;
        unordered_map<string, vector<pair<int, int>>> diagonals;

        // Store all point pairs that can form diagonals in a hash map
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                // Midpoint and square of the diagonal distance
                double midX = (points[i][0] + points[j][0]) / 2.0;
                double midY = (points[i][1] + points[j][1]) / 2.0;
                double distSq = (points[i][0] - points[j][0]) * (points[i][0] - points[j][0]) +
                                (points[i][1] - points[j][1]) * (points[i][1] - points[j][1]);

                string key = to_string(midX) + ":" + to_string(midY) + ":" + to_string(distSq);
                diagonals[key].push_back({i, j});
            }
        }

        // Check for rectangles
        for (auto& entry : diagonals) {
            vector<pair<int, int>>& diagPairs = entry.second;

            for (int i = 0; i < diagPairs.size(); i++) {
                for (int j = i + 1; j < diagPairs.size(); j++) {
                    auto p1 = diagPairs[i].first, p2 = diagPairs[i].second;
                    auto p3 = diagPairs[j].first, p4 = diagPairs[j].second;

                    // Compute area of the rectangle formed by the two diagonals
                    double area = 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])) *
                                  sqrt((points[p2][0] - points[p3][0]) * (points[p2][0] - points[p3][0]) +
                                       (points[p2][1] - points[p3][1]) * (points[p2][1] - points[p3][1]));
                    if (area > 0) {
                        ans = min(ans, area);
                    }
                }
            }
        }

        return ans == (double)INT_MAX ? 0 : ans;
    }
};

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