<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 dsu{
    private:
        int k;
        vector<vector<int>>graph;
        vector<int>par;
        vector<int>rank;
        map<vector<int>,int>mp;
    public:

        dsu(int n,vector<vector<int>>&stones){
            k=n;
            graph=stones;
            par.resize(n,0);
            rank.resize(n,0);
            for(int i=0;i<n;i++)par[i]=i;

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

        int get_parent(int x){
            if(par[x]==x)return x;
            else return par[x]=get_parent(par[x]);
        }

        void union_karo(int i,int j){
            int x1=graph[i][0];
            int y1=graph[i][1];
            int x2=graph[j][0];
            int y2=graph[j][1];

            if(x1==x2 || y1==y2){
                 int a=mp[{graph[i][0],graph[i][1]}];
                 int b=mp[{graph[j][0],graph[j][1]}];

                 a=get_parent(a);
                 b=get_parent(b);

                 if(a!=b){
                    if(rank[a]>rank[b]){
                        par[b]=a;
                    }
                    else if(rank[a]<rank[b]){
                        par[a]=b;
                    }
                    else{
                        par[a]=b;
                        rank[b]++;
                    }
                 }
            }
        }

    int groups(){
        int cnt=0;

        for(int i=0;i<k;i++){
            if(par[i]==i){
                cnt++;
            }
        }

        return cnt;
    }

};


class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        dsu DSU(stones.size(),stones);

        ios_base::sync_with_stdio(false);

        for(int i=0;i<stones.size();i++){
            for(int j=i+1;j<stones.size();j++){
                DSU.union_karo(i,j);
            }
        }

        return stones.size()-(DSU.groups());
    }
};
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