<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 fp(int i, vector<int> & p){

    if(p[i]==i) return i;

    return p[i]= fp(p[i],p);
}

void uni(int i, int j, vector<int> & p, vector<int> & r){

       int  u= fp(i,p);
       int v= fp(j,p);

       if(u==v) return;

       if(r[u]> r[v]) p[v]=u;
       else if(r[v]> r[u]) p[u]=v;

       else{

         p[v]=u;
         r[u]++;
       }
}

    int removeStones(vector<vector<int>>& stones) {

        int maxr=0, maxc=0;

        int n= stones.size();

        for(auto t: stones){

              maxr= max(maxr, t[0]);
              maxc= max(maxc,t[1]);
        }

        vector<int> p(maxr+maxc+2), r(maxr+maxc+2,0);
        for(int i=0; i<maxr+maxc+2; i++) p[i]=i;

        map<int,int> mp;

        for(auto t: stones){


              uni(t[0],t[1]+maxr+1,p,r);

              mp[t[0]]=1;
              mp[t[1]+maxr+1]=1;
        }

        int cnt=0;

         for(auto t: mp){

              if(fp(t.first,p)==t.first) cnt++;
         }


      return n-cnt;



        
    }
};
24626