<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 DisjointSet{
public:
    vector<int> parent,Size;
    DisjointSet(int n){
        parent.resize(n+1);
        Size.resize(n+1,1);

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

    int findUltPar(int u){
        if(parent[u]==u) return u;
        return parent[u]=findUltPar(parent[u]);
    }

    void UnionBySize(int u, int v){
        int ulp_u = findUltPar(u);
        int ulp_v = findUltPar(v);

        if(ulp_u==ulp_v) return ;

        int sizeUp= Size[ulp_u];
        int sizeVp= Size[ulp_v];

        if(sizeUp<sizeVp){
            parent[ulp_u]=ulp_v;
            Size[ulp_v]+=Size[ulp_u];
        }else{
            parent[ulp_v]=ulp_u;
            Size[ulp_u]+=Size[ulp_v];
        }
    }
};

class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        vector<int>row[10001];
        DisjointSet ds(10001);
        sort(stones.begin(),stones.end());
        for(int i=0; i<stones.size(); i++){
            if(i>0 && stones[i]==stones[i-1])
                continue;
            row[stones[i][0]].push_back(stones[i][1]);
        }
        unordered_map<int,int>mp;
        for(int i=0; i<10001; i++){
            for(auto it: row[i]){
                if(mp.find(it)==mp.end()){
                    mp[it]=i;
                }else{
                    ds.UnionBySize(mp[it],i);
                }
            }
        }

        unordered_set<int>ss;
        for(int i=0; i<10001; i++){
            if(row[i].size()==0)continue;
            ss.insert(ds.findUltPar(ds.parent[i]));
        }

        return stones.size()-ss.size();
    }
};
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