<p>On a 2D plane, we place <code>n</code> stones at some integer coordinate points. Each coordinate point may have at most one stone.</p>

<p>A stone can be removed if it shares either <strong>the same row or the same column</strong> as another stone that has not been removed.</p>

<p>Given an array <code>stones</code> of length <code>n</code> where <code>stones[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> represents the location of the <code>i<sup>th</sup></code> stone, return <em>the largest possible number of stones that can be removed</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
<strong>Output:</strong> 5
<strong>Explanation:</strong> One way to remove 5 stones is as follows:
1. Remove stone [2,2] because it shares the same row as [2,1].
2. Remove stone [2,1] because it shares the same column as [0,1].
3. Remove stone [1,2] because it shares the same row as [1,0].
4. Remove stone [1,0] because it shares the same column as [0,0].
5. Remove stone [0,1] because it shares the same row as [0,0].
Stone [0,0] cannot be removed since it does not share a row/column with another stone still on the plane.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> One way to make 3 moves is as follows:
1. Remove stone [2,2] because it shares the same row as [2,0].
2. Remove stone [2,0] because it shares the same column as [0,0].
3. Remove stone [0,2] because it shares the same row as [0,0].
Stones [0,0] and [1,1] cannot be removed since they do not share a row/column with another stone still on the plane.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> stones = [[0,0]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> [0,0] is the only stone on the plane, so you cannot remove it.
</pre>

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

<ul>
	<li><code>1 &lt;= stones.length &lt;= 1000</code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
	<li>No two stones are at the same coordinate point.</li>
</ul>

class Solution {
public:

    void dfs( int i , int j , vector <vector <int>> &stone , map < pair < int , int > , vector < pair <int , int>>> &adj , map <pair <int , int> , int> &vis)
    {
        vis[{i , j}] = 1;

        vector < pair <int , int>> curr = adj[{i , j}];

        for( auto k : curr)
        {
            if( vis[{k.first , k.second}] != 1)
            {
                dfs( k.first , k.second , stone , adj , vis);
            }
        }
    }

    int removeStones(vector<vector<int>>& stones) {
        
       map < pair < int , int > , vector < pair <int , int>>> adj;
       map <pair <int , int> , int> vis;

       int n = stones.size();

       for(int i = 0 ; i < n - 1; i ++ )
       {
            for( int j = i + 1 ; j < n ; j ++)
            {
                if( stones[i][0] == stones[j][0] || stones[i][1] == stones[j][1] ) {
                    adj[{stones[i][0] , stones[i][1]}].push_back({ stones[j][0] , stones[j][1] });
                    adj[{stones[j][0] , stones[j][1]}].push_back({ stones[i][0] , stones[i][1] });
                }
            }
       }

    //    for( auto k : adj)
    //    {
    //         cout << k.first.first << " & " << k.first.second << " - > ";
    //         for( auto p : k.second)
    //         {
    //             cout << "[" << p.first << " " << p.second << "]";

    //         }
    //         cout << endl;
    //    }

       int cnt = 0;
       for( int i = 0 ; i < n ; i ++)
       {
            if( vis[{stones[i][0] , stones[i][1]}] != 1)
            {
                cnt++;
                // cout << stones[i][0] << " " <<  stones[i][1] << endl;
                dfs( stones[i][0] , stones[i][1] , stones , adj , vis);
            }
       }


       return n - cnt;
        
    }
};
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