<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) {
        map<pair<int,int>,bool>mp;
        
        sort(points.begin(),points.end());
        for(auto v:points)
            mp[{v[0],v[1]}]=true;
        int n=points.size();
        int mn=INT_MAX;
        for(int i=0;i<n;i++)
        {
            vector<pair<int,int>>r;
            //look right
            for(int j=i+1;j<n;j++)
            {
                if(points[j][1] == points[i][1])
                    r.push_back({points[j][0],points[j][1]});
            }
            //look down
            for(int j=i-1;j>=0;j--)
            {
                if(points[j][0] == points[i][0])
                {
                    for(auto p:r)
                    {
                        if(mp[{p.first,points[j][1]}])
                            mn=min(mn,(points[i][1]-points[j][1])*(p.first-points[i][0]));
                    }
                    
                }
                else
                    break;
            }
        }
        
        return mn==INT_MAX? 0:mn;
    }
};
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