<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:

    int find(int u,vector<int>& parent){
        if(parent[u]==u)
            return u;
        return parent[u] = find(parent[u],parent);
    }

    void unionSet(int u,int v,vector<int>& parent, vector<int>& rank){
        int pu = find(u,parent);
        int pv = find(v,parent);
        if(rank[pu]<rank[pv])
            parent[pu] = pv;
        else if(rank[pu]>rank[pv])
            parent[pv] = pu;
        else{
            parent[pv] = pu;
            rank[pu]++;
        }
    }

    int removeStones(vector<vector<int>>& stones) {
        int maxrow = 0;
        int maxcol = 0;
        for(auto i:stones){
            maxrow = max(maxrow,i[0]+1);
            maxcol = max(maxcol,i[1]+1);
        }
        int n = maxrow+maxcol;
        cout<<maxrow<<" "<<maxcol<<" "<<n<<endl;
        vector<int> parent(n);
        vector<int> rank(n);
        for(int i=0;i<n;i++){
            parent[i] = i;
            rank[i] = 0;
        }
        
        for(auto i:stones){
            int r = i[0];
            int c = i[1]+maxrow;
            unionSet(r,c,parent,rank);
            // m[r] = 1;
            // m[c] = 1;
        }
        unordered_map<int,int> m;
        int stay=0;
        for(auto i:parent){
            m[find(i,parent)]++;
            if(m[find(i,parent)]==2){
                stay++;
                cout<<i<<endl;
            }
        }
        return stones.size()-stay;
    }
};
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