<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 {
    vector<int> root, Size;
public:
    int merge;
    unionFind(int N) : root(N), Size(N, 1), merge(0) {
        iota(root.begin(), root.end(), 0);
    }

    int Find(int x) {
        if (x == root[x]) return x;
        return root[x] = Find(root[x]); // Path compression
    }

    bool Union(int x, int y) {
        x = Find(x), y = Find(y);

        if (x == y) return 0;

        if (Size[x] > Size[y]) {
            Size[x] += Size[y];
            root[y] = x;
        } 
        else {
            Size[y] += Size[x];
            root[x] = y;
        }
        merge++;
        return 1;
    }
};

class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        int sz = stones.size();
        int M=10001, N=2*M+1;
        unionFind G(N);
        bitset<20003> cntRC=0;
        for (int idx = 0; idx < sz; idx++) {
            const int i = stones[idx][0], j=stones[idx][1];
            G.Union(i, M+j);
            cntRC[i]=cntRC[M+j]=1;
        }
        return sz-cntRC.count()+G.merge;
    }
};
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