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

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

        void unionsize(int u,int v){
            int upar=parent(u);
            int vpar=parent(v);
            if(vpar==upar)return;
            else if(size[vpar]>size[upar]){
                size[vpar]+=size[upar];   
                par[upar]=vpar;
            }
            else{
             size[upar]+=size[vpar];   
                par[vpar]=upar;   
            }
        }
        int getPar(int i) {
        return par[i];
    }

    
    int getSize(int i) {
        return size[i];
    }

    };
public:
    int removeStones(vector<vector<int>>& stones) {
        int n=stones.size();
        int maxRow=0;
        int maxCol=0;
        for(auto it :stones){
            maxRow=max(maxRow,it[0]);
            maxCol=max(maxCol,it[1]);
        }
        disjoint ds(maxRow + maxCol + 1);
        for(int i=0;i<n;i++){
            int u=stones[i][0];
            int v=stones[i][1]+ 1 + maxRow;
            ds.unionsize(u,v);
        }
         int c = maxRow + maxCol + 2;
        for(int i=0;i<c;i++){
            if(ds.getPar(i)==i  && ds.getSize(i)>1)
            n--;
    }
    return n;
    }
};
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