<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 makeSet(auto& rank,auto& vals,int n){
        for(int i=0;i<n;i++){
            // p[{val[i][0],val[i][1]}]={val[i][0],val[i][1]};
            rank[{vals[i][0],vals[i][1]}]=0;
        }
    }
    pair<int,int> find(auto u,auto& p){
        if(p[u] == u) return u;

        return p[u]= find(p[u],p);
    }
    
    void uni(auto u, auto v, auto& p, auto& rank){
        u = find(u,p);
        v = find(u,p);

        if(rank[u]<rank[v]){
            p[u]=v;
        }else if(rank[v]< rank[u]){
            p[v]=u;
        }else{
            p[v]=u;
            rank[u]++;
        }
    }
    int removeStones(vector<vector<int>>& stones) {
        int n= stones.size();
        map<pair<int,int>,vector<pair<int,int>>> adj;
        map<pair<int,int>,int> vis;
        makeSet(vis,stones,n);

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

        queue<pair<int,int>> q;
        int ans =0;

        for(int i=0;i<n;i++){
            if(vis[{stones[i][0],stones[i][1]}]) continue;

            q.push({stones[i][0],stones[i][1]});
            vis[{stones[i][0],stones[i][1]}]=1;
            ans++;
            while(!q.empty()){
                auto node = q.front();
                q.pop();

                for(auto it:adj[node]){
                    if(!vis[it]){
                        vis[it]=1;
                        q.push(it);
                    }
                }
            }
        }

        if(n==1) return 0;

        return n-ans;      
    }
};
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