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

    int find(int n){
        if(parent[n]!=n) parent[n]=find(parent[n]);
        return parent[n];
    }

    void union_set(int x,int y){
        int xset = find(x),yset=find(y);
        if(yset==xset){
            return;
        }
        else if(rank[xset]>rank[yset]){
            rank[xset]+=rank[yset];
            parent[yset]=xset;
        }
        else{
            rank[xset]+=rank[yset];
            parent[yset]=xset;
        }
    }
};

class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        unordered_map<int,vector<int>> col;
        unordered_map<int,vector<int>> row;

        int j=0;
        for(auto stone:stones){
            col[stone[1]].push_back(j);
            row[stone[0]].push_back(j);
            j++;
        }

        UnionFind dsu(stones.size());

        for(auto c:col){
            int x=c.second[0];
            for(int i=1;i<c.second.size();i++){
                dsu.union_set(x,c.second[i]);
            }
        }
        for(auto c:row){
            int x=c.second[0];
            for(int i=1;i<c.second.size();i++){
                dsu.union_set(x,c.second[i]);
            }
        }
        int ans=0;
        for(int i=0;i<stones.size();i++){
            if(dsu.parent[i]==i) ans++;
        }
        return stones.size()-ans;
    }
};
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