Minimum Area Rectangle
<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 a rectangle formed from these points, with sides parallel to the X and Y axes</em>. If there is not any such rectangle, return <code>0</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0900-0999/0939.Minimum%20Area%20Rectangle/images/rec1.jpg" style="width: 500px; height: 447px;" />
<pre>
<strong>Input:</strong> points = [[1,1],[1,3],[3,1],[3,3],[2,2]]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0900-0999/0939.Minimum%20Area%20Rectangle/images/rec2.jpg" style="width: 500px; height: 477px;" />
<pre>
<strong>Input:</strong> points = [[1,1],[1,3],[3,1],[3,3],[4,1],[4,3]]
<strong>Output:</strong> 2
</pre>

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

<ul>
	<li><code>1 &lt;= points.length &lt;= 500</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>
Geometry; Array; Hash Table; Math; Sorting
class Solution {
public:
    int minAreaRect(vector<vector<int>>& points) {
        map<int, vector<int>> d;
        for (auto& p : points) {
            int x = p[0], y = p[1];
            d[x].emplace_back(y);
        }
        unordered_map<int, int> pos;
        int ans = 1 << 30;
        for (auto& [x, ys] : d) {
            sort(ys.begin(), ys.end());
            int n = ys.size();
            for (int i = 0; i < n; ++i) {
                int y1 = ys[i];
                for (int j = i + 1; j < n; ++j) {
                    int y2 = ys[j];
                    int p = y1 * 40001 + y2;
                    if (pos.count(p)) {
                        ans = min(ans, (x - pos[p]) * (y2 - y1));
                    }
                    pos[p] = x;
                }
            }
        }
        return ans == 1 << 30 ? 0 : ans;
    }
};