<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,rank,size;
    DisjointSet(int n) {
        parent.resize(n);
        rank.resize(n,0);
        size.resize(n,1);
        for(int i=0;i<n;++i) parent[i] = i;
    }
    int getparent(int u) {
        if(parent[u]==u) return u;
        int pu = getparent(parent[u]);
        parent[u] = pu;   // path compression laga diya bhai
        return pu;
    }
    void unionbyrank(int u,int v) {
        int pu = getparent(u);
        int pv = getparent(v);
        if(pu==pv) return;

        if(rank[pu]>rank[pv]) {
            rank[pu] = max(rank[pu],rank[pv]+1);
            parent[pv] = pu;
        }
        else {
            rank[pv] = max(rank[pv],rank[pu]+1);
            parent[pu] = pv;
        }
    }
};
    int removeStones(vector<vector<int>>& stones) {
        int n = stones.size();
        int maxrow = 0,maxcol = 0;
        for(auto it:stones) {
            maxrow = max(maxrow,it[0]);
            maxcol = max(maxcol,it[1]);
        }
        DisjointSet ds(maxrow+maxcol+2);
        unordered_map<int,int> stoneNodes;
        for(int i=0;i<n;++i) {
            int row = stones[i][0];
            int col = stones[i][1]+maxrow+1;
            if(ds.getparent(row)!=ds.getparent(col)) {
                ds.unionbyrank(row,col);
            }
            stoneNodes[row] = 1;
            stoneNodes[col] = 1;
        }
        int cnt = 0;
        for(auto it:stoneNodes) {
            if(ds.getparent(it.first)==it.first) {
                cnt++;
            }
        }

        return n-cnt;
    }
};
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