<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 {
    vector<int>parent,Rank;
public:
    int removeStones(vector<vector<int>>& stones) {
        int n = stones.size();
        if (n <= 1) return 0;
        int maxrow = 0,maxcol = 0;
        for(auto it : stones){
            maxrow = max(maxrow, it[0]);
            maxcol = max(maxcol, it[1]);
        }
        parent.resize(maxrow + maxcol + 2);
        Rank.resize(maxrow + maxcol + 2, 0);
        for(int i = 0; i < maxrow + maxcol + 2; i++){
            parent[i] = i;
        }
        unordered_map<int,int>stone;
        for(auto it : stones){
            int rownode = it[0];
            int colnode = it[1] + maxrow + 1;
            unionSet(rownode,colnode);
            stone[rownode] = 1;
            stone[colnode] = 1;
        }
        int cnt = 0;
        for(auto it : stone){
            if(find(it.first) == it.first)cnt++;
        }
        return (n - cnt);
    }
    int find(int X)
    {
        if (X != parent[X]) {
            parent[X] = find(parent[X]); 
        }
        return parent[X];
    }
    void unionSet(int X,int Z)
    {
        int up_X = find(X);
        int up_Z = find(Z);
        if(up_X == up_Z) return;
        if(Rank[up_X] < Rank[up_Z]){
            parent[up_X] = up_Z;
        }
        else if(Rank[up_Z] < Rank[up_X]){
            parent[up_Z] = up_X;
        }
        else {
            parent[up_Z] = up_X;
            Rank[up_X]++;
        }
    }
};
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