<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 {
    static constexpr int n = 10001;

    class DisjointSet {
        struct Node {
            int parent = -1;
            int rank = 1;
        };
        mutable vector<Node> nodes;
        int numSets = 0;
    public:
        DisjointSet(int size) : nodes(size) {}

        void add(int n) {
            if(nodes[n].parent == -1) {
                nodes[n].parent = n;
                numSets++;
            }
        }

        int find(int n) const {
            if(nodes[n].parent == n)
                return n;
            else if(nodes[n].parent == -1)
                return -1;
            return nodes[n].parent = find(nodes[n].parent);
        }

        bool merge(int n1, int n2) {
            const int rn1 = find(n1);
            const int rn2 = find(n2);
            if(rn1 == rn2)
                return false;
            
            numSets--;
            if(nodes[rn1].rank >= nodes[rn2].rank) {
                nodes[rn2].parent = rn1;
                if(nodes[rn1].rank == nodes[rn2].rank)
                    nodes[rn1].rank++;
            } else {
                nodes[rn1].parent = rn2;
            }
            return true;
        }

        int getNumSets() const { return numSets; }
    };
public:
    int removeStones(vector<vector<int>>& stones) {
        const int ss = stones.size();
        DisjointSet djs(n * 2);

        for(const auto& s : stones) {
            djs.add(s[0]);
            djs.add(s[1] + n);
            djs.merge(s[0], s[1] + n);
        }
        return ss - djs.getNumSets();
    }
};
30648