<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:
    int removeStones(vector<vector<int>>& stones) {
        int n = stones.size();
        parent.resize(n);
        rank.resize(n, 1);

        // initialize
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }

        unordered_map<int, vector<int>> row_map, col_map;


        for (int i = 0; i < n; i++) {
            row_map[stones[i][0]].push_back(i);
            col_map[stones[i][1]].push_back(i);
        }

        for (auto& row : row_map) {
            for (int i = 1; i < row.second.size(); i++) {
                Union(row.second[0], row.second[i]);
            }
        }

        for (auto& col : col_map) {
            for (int i = 1; i < col.second.size(); i++) {
                Union(col.second[0], col.second[i]);
            }
        }

        unordered_set<int> seen;
        for (int i = 0; i < n; i++) {
            seen.insert(Find(i));
        }

        return n - seen.size();

    }
private:
    vector<int> parent, rank;

    int Find(int x) {
        return x == parent[x] ? x : parent[x] = Find(parent[x]);
    }

    void Union(int x, int y) {
        int rootX = Find(x);
        int rootY = Find(y);
        if (rank[rootX] > rank[rootY]) {
            parent[rootY] = rootX;  
        } else if (rank[rootX] < rank[rootY]) {
            parent[rootX] = rootY; 
        } else {
            parent[rootY] = rootX;
            rank[rootX]++;
        }
    }
};
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