<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> par;
    vector<int> rank;
    vector<int> size;
    Disjoint(int n){
        par.resize(n+1);
        rank.resize(n+1,0);
        size.resize(n+1,1);
        for(int i=0;i<=n;i++){
            par[i]=i;
        }
    }
    int findpar(int node){
        if(par[node]==node)return node;
        return par[node]=findpar(par[node]);
    } 
    void unionrank(int a,int b){
        int pa=findpar(a);
        int pb=findpar(b);
        if(pa==pb)return;
        if(rank[pa]>rank[pb]){
            par[pb]=pa;
        }else if(rank[pa]==rank[pb]){
            par[pb]=pa;
            rank[pa]++;
        }else{
            par[pa]=pb;
        }     
    } 
    void unionsize(int a,int b){
        int pa=findpar(a);
        int pb=findpar(b);
        if(pa==pb)return;
        if(size[pa]>=size[pb]){
            par[pb]=pa;
            size[pa]+=size[pb];
        }else{
            par[pa]=pb;
            size[pb]+=size[pa];
        }
    } 
};


class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        int n=stones.size();
        int r=0;
        int c=0;
        for(int i=0;i<n;i++){
            r=max(r,stones[i][0]);
            c=max(c,stones[i][1]);
        }
        
        Disjoint D(r+1+c+1);
        for(int i=0;i<n;i++){
            D.unionsize(stones[i][0],r+1+stones[i][1]);
        }
        int g=0;
        unordered_map<int,int> mp;
        for(int i=0;i<r+c+2;i++){
            int p=D.findpar(i);
            
            if(mp[p]==1)continue;
            
            mp[p]=1;
            if(D.size[p]==1)continue;
            // cout<<i<<" "<<p<<endl;
            g++;
        }
        return n-g;
        
        
        
    }
};
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