<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 DisjointSet {
public:
    vector<int> parent, rank, size;
    DisjointSet(int n) {
        rank.resize(n + 1, 0);
        size.resize(n + 1, 1);
        parent.resize(n + 1);
        for (int i = 0; i <= n; i++) {
            parent[i] = i;
        }
    }
    int findUpar(int u) {
        if (parent[u] == u) {
            return u;
        }
        return parent[u] = findUpar(parent[u]);
    }
    void UnionByRank(int u, int v) {
        int up = findUpar(u);
        int vp = findUpar(v);
        if (up == vp) return;
        if (rank[up] < rank[vp]) {
            parent[up] = vp;
        } else if (rank[up] > rank[vp]) {
            parent[vp] = up;
        } else {
            parent[vp] = up;
            rank[up]++;
        }
    }
    void UnionBySize(int u, int v) {
        int up = findUpar(u);
        int vp = findUpar(v);
        if (up == vp) return;
        if (size[up] < size[vp]) {
            parent[up] = vp;
            size[vp] += size[up];
        } else {
            parent[vp] = up;
            size[up] += size[vp];
        }
    }
};
class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        int n = stones.size();
        int maxRow = 0, maxCol = 0;
        for (auto& stone : stones) {
            maxRow = max(maxRow, stone[0]);
            maxCol = max(maxCol, stone[1]);
        }
        DisjointSet ds(maxRow + maxCol + 1);
        for (auto& stone : stones) {
            int rowNode = stone[0];
            int colNode = stone[1] + maxRow + 1;
            ds.UnionBySize(rowNode, colNode);
        }
        unordered_set<int> uniqueParents;
        for (auto& stone : stones) {
            int rowNode = stone[0];
            uniqueParents.insert(ds.findUpar(rowNode));
        }
        return n - uniqueParents.size();
    }
};

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