<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 {
public:
vector<int> parent;
vector<int> size;
unordered_map<int, vector<int>> map;
unordered_set<int> set;
int n;
    int removeStones(vector<vector<int>>& stones) {
        n = stones.size();

        parent.resize(20001, 20001);

        for (int i = 0; i < parent.size(); i++)
            parent[i] = i;

        size.resize(20001, 1);

        for (auto& stone : stones) {
            int row = stone[0];
            int col = stone[1] + 10001; // Offset columns by 10001 to differentiate from rows

            if (find(row) != find(col)) {
                unionMethod(row, col);
            }
        }

        for (auto& stone : stones) {
            set.insert(find(stone[0]));  // You can use either row or col
        }

        return stones.size() - set.size();
        
    }

    int find (int index) {
        if (index == parent[index])
            return index;

        return parent[index] = find(parent[index]);
    }

    bool unionMethod(int x, int y) {
        int x_parent = find(x);
        int y_parent = find(y);

        if (x_parent == y_parent)
            return true;

        if (size[x_parent] == size[y_parent]) {
            parent[y_parent] = x_parent;
            size[x_parent] += size[y_parent];
        } else if (size[x_parent] < size[y_parent])
            parent[x_parent] = y_parent;
        else if (size[x_parent] > size[y_parent])
            parent[y_parent] = x_parent;

        return true;
    }
};

/*
[[0,0]]
*/
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