<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 DSU{
    private:
        vector<int> parent;
        vector<int> size;
        
    public:
        DSU(int n) {
            parent.resize(n);
            size.resize(n, 1);
            for(int i = 0; i < n; i++)
                parent[i] = i;
        }
        
        int findParent(int node) {
            if (parent[node] == node)
                return node;
            return parent[node] = findParent(parent[node]);  // Path compression
        }
        
        void unionBySize(int node1, int node2) {
            int parent1 = findParent(node1);
            int parent2 = findParent(node2);
            if (parent1 != parent2) {
                if (size[parent1] < size[parent2]) {
                    parent[parent1] = parent2;
                    size[parent2] += size[parent1];
                } else {
                    parent[parent2] = parent1;
                    size[parent1] += size[parent2];
                }
            }
        }
    };
    
    int removeStones(vector<vector<int>>& stones) {
        int n = stones.size();
        int maxRow = -1;
        int maxCol = -1;
        
        // Find the maximum row and column indices
        for (auto& stone : stones) {
            maxRow = max(maxRow, stone[0]);
            maxCol = max(maxCol, stone[1]);
        }
        
        // DSU for rows and columns combined, where row indices and column indices are separate
        DSU dsu(maxRow + maxCol + 2);
        
        // Union rows and columns for each stone
        for (auto& stone : stones) {
            int row = stone[0];
            int col = stone[1] + maxRow + 1;  // Ensure column indices are distinct from row indices
            dsu.unionBySize(row, col);
        }
        
        // Find how many unique connected components there are
        set<int> uniqueParents;
        for (auto& stone : stones) {
            int row = stone[0];
            uniqueParents.insert(dsu.findParent(row));  // Only count row or column once
        }
        
        // The number of moves is the total number of stones minus the number of connected components
        return stones.size() - uniqueParents.size();
    }
};

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