<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:
    pair<int,int> findparent(pair<int,int> &a, map<pair<int,int>,pair<int,int>> &parent){
        if(parent[a]==a) return a;
        parent[a]=findparent(parent[a],parent);
        return parent[a];
    }
    void unionn(pair<int,int> &a, pair<int,int> &b, 
    map<pair<int,int>,pair<int,int>> &parent, map<pair<int,int>,int> &rank){
        pair<int,int> up = findparent(a,parent);
        pair<int,int> vp = findparent(b,parent);
        if(rank[up]>rank[vp]){
            parent[vp]=up;
        }
        else if(rank[up]<rank[vp]){
            parent[up]=vp;
        }
        else{
            rank[up]++;
            parent[vp]=up;
        }
        return;
    }
    int removeStones(vector<vector<int>>& stones) {
        int n = stones.size();
        map<pair<int,int>,pair<int,int>> parent;
        map<pair<int,int>,vector<pair<int,int>>> adjlist;
        map<pair<int,int>,int> rank;
        for(int i=0;i<n;i++){
            int s = stones[i][0];
            int e = stones[i][1];
            parent[{s,e}]={s,e};
            for(int j=0;j<n;j++){
                if((stones[j][0]==s && stones[j][1]!=e)||
                ((stones[j][1]==e && stones[j][0]!=s))){
                    adjlist[{s,e}].push_back({stones[j][0],stones[j][1]});
                }
            }
        }
        for(int i=0;i<n;i++){
            pair<int,int> t = {stones[i][0],stones[i][1]};
            for(auto it : adjlist[t]){
                unionn(it,t,parent,rank);
            }
        }
        int count = 0;
        for(auto it : parent){
            if((it.first.first == it.second.first)&&(it.first.second == it.second.second)) count++;
        }
        
        return n-count;
    }
};
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