<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 {
    unordered_map<int, int> Father;
    unordered_map<int, vector<int>> col;
    unordered_map<int, vector<int>> row;
    int N = 10000;  //stones length = 1000
public:
    int removeStones(vector<vector<int>>& stones)
    {
        for (auto x : stones)
        {
            int i = x[0];
            int j = x[1];
            int id = i * N + j;
            Father[id] = id;
            col[i].push_back(id);
            row[j].push_back(id);
        }

        for (auto c : col)
        {
            int id0 = c.second[0];
            for (int i = 1; i < c.second.size(); ++i)
            {
                int id = c.second[i];
                if (Find(id0) != Find(id))
                    Union(id0, id);                
            }
        }
        for(auto r : row)
        {
            int id0 = r.second[0];
            for(int i = 1; i < r.second.size(); ++i)
            {
                int id = r.second[i];
                if(Find(id0) != Find(id))
                    Union(id0, id);
            }
        }

        unordered_set<int> Set;
        for(auto s : stones)
        {
            int id = s[0] * N + s[1];
            int id0 = Find(id);
            Set.insert(id0);
        }
        return stones.size() - Set.size();
    }
    int Find(int x)
    {
        if(Father[x] != x)
            Father[x] = Find(Father[x]);
        return Father[x];
    }
    void Union(int x, int y)
    {
        x = Father[x];
        y = Father[y];
        if(x <= y)
            Father[y] = x;
        else
            Father[x] = y;
    }
};
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