<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://assets.leetcode.com/uploads/2021/08/03/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://assets.leetcode.com/uploads/2021/08/03/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>

class Solution {
public:
    int minAreaRect(vector<vector<int>>& points) {
        unordered_map<int, unordered_set<int>> x;
        //unordered_map<int, unordered_set<int>> y;
        int ans = INT_MAX;

        for (auto& point: points) {
            x[point[0]].insert(point[1]);
            //y[point[1]].insert(point[0]);
        }

        for (auto iter = x.begin(); iter != x.end(); iter++) {
            for (auto iter1 = next(iter); iter1 != x.end(); iter1++) {
                set<int> yy;
                for (int it: iter->second) {
                    if (iter1->second.count(it)) {
                        yy.insert(it);
                    }
                }

                for (auto iter2 = yy.begin(); iter2 != yy.end() && next(iter2) != yy.end(); iter2++) {
                    ans = min(ans, abs(iter->first - iter1->first)*abs(*iter2 - *(next(iter2))));
                }
            }
        }

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