<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:

class dsu{
	vector<int> parent, size, rank;

public:
	dsu(int n){
		for(int i=0; i<=n; i++){
			size.push_back(1);
			parent.push_back(i);
			rank.push_back(0);
		}
	}

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

	void unionBysize(int u, int v){
		int pu= findpar(u);
		int pv= findpar(v);

		if(pu==pv){
			return;
		}
		if(size[pu]<size[pv]){
			parent[pu]= pv;
			size[pv]+= size[pu];
		}else{
			parent[pv]= pu;
			size[pu]+= size[pv];
		}
	}
};


    int removeStones(vector<vector<int>>& stones) {
        int n=0;
        for(auto it:stones){
            for(auto itr: it){
                n=max(n, itr);
            }
        }
        n+=2;
        int k=n;
        n*=2;

        dsu d(n);
        set<int> s;
        for(auto it: stones){
            int u= it[0], v= it[1];
            v+=k;
            s.insert(u);
            s.insert(v);
            d.unionBysize(u,v);
        }

        int x=0;
        for(auto it: s){
            if(d.findpar(it)==it){
                x++;
            }
        }

        int sz= stones.size();
        return sz-x;
    }
};
30648