<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>

#define ll long long
class DisjointSet {
public:
    vector<ll> rank, size, parent;
    DisjointSet(int n) {
        rank.resize(n + 1, 0);
        size.resize(n + 1, 1);
        parent.resize(n + 1, 1);
        for (int i = 0; i < n + 1; i++)
            parent[i] = i;
    }

    int fup(int node) {
        if (parent[node] == node)
            return node;
        return parent[node] = fup(parent[node]);
    }

    void ubr(int a, int b) {
        int pa = fup(a), pb = fup(b);
        if (pa == pb)
            return;
        if (rank[pa] > rank[pb])
            parent[pb] = pa;
        else if (rank[pa] < rank[pb])
            parent[pa] = pb;
        else {
            parent[pa] = pb;
            rank[pb]++;
        }
    }
    void ubs(int a, int b) {
        int pa = fup(a), pb = fup(b);

        if (size[pa] > size[pb])
            parent[pb] = pa, size[pa] += size[pb];
        else
            parent[pa] = pb, size[pb] += size[pa];
    }
};

class Solution {
public:
    int removeStones(vector<vector<int>>& arr) {
        int n = arr.size();
        DisjointSet ds(n);

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int x = arr[i][0], y = arr[i][1];
                int xx = arr[j][0], yy = arr[j][1];

                if (x == xx || y == yy) {
                    if (ds.fup(i) != ds.fup(j))
                        ds.ubr(i, j);
                }
            }
        }

        int cnt{};
        for (int i = 0; i < n; i++)
            if (ds.parent[i] == i)
                cnt++;

        return n - cnt;
    }
};
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