<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:
    class DisjointSet {
    public:
        vector<int> parent, size;

        DisjointSet(int n) {
            parent.resize(n + 1);
            size.resize(n + 1, 1);
            for (int i = 0; i <= n; i++) {
                parent[i] = i;
            }
        }

        int findUParent(int u) {
            if (parent[u] == u) {
                return u;
            }
            return parent[u] = findUParent(parent[u]);  // Path compression
        }

        void UnionBySize(int u, int v) {
            int ulpu = findUParent(u);
            int ulpv = findUParent(v);

            if (ulpu != ulpv) {
                if (size[ulpu] < size[ulpv]) {
                    parent[ulpu] = ulpv;
                    size[ulpv] += size[ulpu];
                } else {
                    parent[ulpv] = ulpu;
                    size[ulpu] += size[ulpv];
                }
            }
        }
    };

    int removeStones(vector<vector<int>>& stones) {
        int n = stones.size(); // Initialize number of stones
        int maxrow = 0;
        int maxcol = 0;
        
        // Find max row and column
        for (auto& it : stones) {
            maxrow = max(maxrow, it[0]);
            maxcol = max(maxcol, it[1]);
        }

        // Create DisjointSet with size maxrow + maxcol + 1
        DisjointSet ds(maxrow + maxcol + 1);
        unordered_map<int, int> stonenodes;

        // Union rows and columns
        for (auto& it : stones) {
            int noderow = it[0];
            int nodecol = it[1] + maxrow + 1; // Offset column to differentiate from row
            ds.UnionBySize(noderow, nodecol);
            stonenodes[noderow] = 1;  // Track nodes from row
            stonenodes[nodecol] = 1;  // Track nodes from column (with offset)
        }

        // Count number of connected components
        int cnt = 0;
        for (auto& it : stonenodes) {
            if (ds.findUParent(it.first) == it.first) {
                cnt++;
            }
        }

        // The answer is the number of stones minus the number of components
        return n - cnt;
    }
};

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