<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>>& a) {
        auto foo = ios_base::sync_with_stdio (false);
        int i,j,n=a.size();
        long long ans = 1e12;

        unordered_map<int, unordered_set<int> >m;
        for(auto p:a)m[p[0]].insert(p[1]);

        for(i=0;i<n;i++)
        {
            for(j=i+1;j<n;j++)
            {
                if(a[i][0] == a[j][0] || a[i][1] == a[j][1])continue;

                // it can be a diagonal
                vector<int>pntx = {a[i][0], a[j][1]};
                vector<int>pnty = {a[j][0], a[i][1]};

                if((m.find(pntx[0])!=m.end() && m[pntx[0]].find(pntx[1])!=m[pntx[0]].end()) 
                && (m.find(pnty[0])!=m.end() && m[pnty[0]].find(pnty[1])!=m[pnty[0]].end()))
                {
                    long long x,y,area;
                    x = abs(a[i][0] - a[j][0]);
                    y = abs(a[i][1] - a[j][1]);
                    area = x*y;
                    ans = min(ans, area);
                }
            }
        }
        if(ans < 0 || ans>=1e12)ans=0;
        return ans;
    }
};
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