<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 Disjoint{
    public:
    vector<int>size,parent;
    Disjoint(int n){
        size.resize(n+1,1);
        parent.resize(n+1);
        for(int i=0;i<=n;i++){
            parent[i]=i;
        }
    }

    
    int findpar(int x){
        if(x==parent[x]) return x;
        return parent[x]=findpar(parent[x]);
    }

    void unionBySize(int u,int v){
        int up=findpar(u);
        int vp=findpar(v);

        if(up==vp) return;
        if(size[up]<size[vp]){
            parent[up]=vp;
            size[vp]+=size[up];
        }
        else{
            parent[vp]=up;
            size[up]+=size[vp];
        }
    }
};



class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        int n=stones.size();
        int max_row=0;
        int max_col=0;

        for(auto it : stones){
            max_row=max(max_row,it[0]);
            max_col=max(max_col,it[1]);
        }

        Disjoint ds(max_row+max_col+1);

        unordered_map<int,int>mpp;

        for(auto it: stones){
            int nrow=it[0];
            int ncol=it[1]+max_row+1;

            ds.unionBySize(nrow,ncol);
            mpp[nrow]=1;
            mpp[ncol]=1;
        }

        int cnt=0;
        for(auto it : mpp){
            if(ds.findpar(it.first)==it.first)  {
                cnt++;
            }
        }


        return n-cnt;

    }
};
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