<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:
    int findpar(int v, vector<int>& par)
    {
        if(par[v]==v)
        {
            return v;
        }
        return par[v]=findpar(par[v],par);
    }
    void unionsize(int x, int y, vector<int>& par, vector<int>& sz)
    {
        int u = findpar(x,par);
        int v = findpar(y,par);
        if(u==v) return;
        if(sz[u]<sz[v])
        {
            par[u] = v;
            sz[v] += sz[u];
        }
        else
        {
            par[v] = u;
            sz[u] += sz[v];
        }
    }
    int removeStones(vector<vector<int>>& stones) 
    {
        int n = stones.size();
        int maxr = 0;
        int maxc = 0;
        for(int i=0; i<n; i++)
        {
            maxr = max(maxr,stones[i][0]);
            maxc = max(maxc,stones[i][1]);
        }
        vector<int> par(maxr+maxc+2);
        vector<int> sz(maxr+maxc+2,1);
        for(int i=0; i<par.size(); i++)
        {
            par[i] = i;
        }
        unordered_set<int> used;
        for(int i=0; i<n; i++)
        {
            int u = stones[i][0];
            int v = stones[i][1]+maxr+1;
            if(findpar(u,par)!=findpar(v,par))
            {
                unionsize(u,v,par,sz);
                used.insert(u);
                used.insert(v);
            }
        }
        int comps=0;
        for(auto x:used)
        {
            // cout<<x<<" "<<findpar(x,par)<<endl;
            if(findpar(x,par)==x)
            {
                comps++;
            }
        }
        int removed = n - comps;
        return removed;
    }
};

























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