<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 DisjointSet{
    
    public:
    vector<int>rank,parent;

    DisjointSet(int n){
        rank.resize(n+1,0);
        parent.resize(n+1);

        for(int i=0; i<=n; i++){
            parent[i]=i;
        }
    }

    int find(int node){
        if(node==parent[node]){
            return node;
        }

        return parent[node]=find(parent[node]);
    }

    void Union(int u,int v){
        int u_parent=find(u);
        int v_parent=find(v);

        if(u_parent==v_parent){
            return ;
        }

        if(rank[u_parent]<rank[v_parent]){
              parent[u_parent]=v_parent;
        }
        else if(rank[u_parent]>rank[v_parent]){
              parent[v_parent]=u_parent;
        }
        else{
             parent[v_parent]=u_parent;
             rank[u_parent]++;
        }
    }    
};

class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        int m=stones.size();
       

        vector<pair<int,int>>edges;
        for(int i=0; i<m; i++){
            for(int j=i+1; j<m; j++){
                if(stones[i][0]==stones[j][0] || stones[i][1]==stones[j][1]){
                     edges.push_back({i,j});
                }
            }
        }
        DisjointSet ds(m);

        for(auto &it:edges){
            int u=it.first;
            int v=it.second;

            if(ds.find(u)==ds.find(v)){
                continue;
            }

            else{
                ds.Union(u,v);
            }
        }
        int count=0;
        for(int i=0; i<m; i++){
            if(i==ds.parent[i]){
                count++;
            }
        }

        cout<<count;

        return m-count;
        


    }
};
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