<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 DSU {
    unordered_map<int, int> parent, rank;
public:
    int count = 0;
    DSU(int n, vector<vector<int>> &stones)
    {
        count = n;
        for(auto &s : stones)
        {
            int cell = s[0]*1e4 + s[1];
            parent[cell] = cell;
            rank[cell] = 0;
        }
    }

    int find(int x)
    {
        if(parent[x] == x)
            return x;
        else
            return parent[x] = find(parent[x]);
    }

    bool union_sets(int x, int y)
    {
        int px = find(x), py = find(y);

        if(px == py)
            return false;

        if(rank[px] > rank[py])
        {
            parent[py] = px;
        }        
        else if(rank[px] < rank[py])
        {
            parent[px] = py;
        }
        else
        {
            parent[py] = px;
            rank[px]++;
        }
        return true;
    }

};

class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        int n = stones.size();
        DSU dsu = DSU(n, stones);

        sort(stones.begin(), stones.end());
        for(int i = 1; i < n; i++)
        {
            if(stones[i][0] == stones[i - 1][0])
            {
                int index1 = stones[i][0] * 1e4 + stones[i][1];
                int index2 = stones[i - 1][0] * 1e4 + stones[i - 1][1];
                if(dsu.union_sets(index1, index2))
                    dsu.count--;
            }
        }

        sort(stones.begin(), stones.end(), [](vector<int> &a, vector<int> &b){
            return a[1] < b[1];
        });
        for(int i = 1; i < n; i++)
        {
            if(stones[i][1] == stones[i - 1][1])
            {
                int index1 = stones[i][0] * 1e4 + stones[i][1];
                int index2 = stones[i - 1][0] * 1e4 + stones[i - 1][1];
                if(dsu.union_sets(index1, index2))
                    dsu.count--;
            }
        }

        int ans = n - dsu.count;
        return ans;
    }
};
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