<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, set<int>> p;

        for(int i=0;i<points.size();i++){
            p[points[i][0]].insert(points[i][1]);
        }

        int ans = -1;

        for(auto&[k1,v1]:  p){
            for(auto&[k2,v2]: p){
                //take intersection
                if(k1 == k2){
                    continue;
                }

                set<int> intersection;
                
                set_intersection(v1.begin(), v1.end(), v2.begin(), v2.end(),  std::inserter(intersection, intersection.begin()));

                if(intersection.size() <= 1){
                    continue;
                }

                for(auto itr = intersection.begin(); next(itr) != intersection.end(); ++itr){
                    auto nxt = next(itr);

                    int val = abs(k1-k2)*(abs(*itr - *nxt));

                    if(ans == -1){
                        ans = val;
                    }
                    else{
                        ans = min(ans, val);
                    }
                }
            }
        }

        return (ans == -1) ? 0 : ans;
    }
};
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