<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 dsu{
    vector<int> parent,size,rank;
    public: 
     dsu(int n){
        for(int i = 0 ; i< n ; i++){
            parent.push_back(i);
            size.push_back(1);
            rank.push_back(0);
        }
     }
     int findParent(int node){
        if(parent[node]== node) return node;
        return parent[node]=findParent(parent[node]);
     }
     void unionSize(int u,int v){
        int pu = findParent(u);
        int pv = findParent(v);
        if(pu==pv) return;

        if(size[pu] < size[pv]){
            parent[pu]=pv;
            size[pv] +=size[pu];
        }else{
            parent[pv]=pu;
            size[pu] +=size[pv];
        }
     }
};
class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        int maxRow = 0;
        int maxCol = 0;
        for(auto it : stones){
            maxRow = max(maxRow,it[0]);
            maxCol = max(maxCol,it[1]);
        }
        dsu du(maxRow + maxCol+2);
        for(auto it : stones){
            int col = it[1]+maxRow+1;
            du.unionSize(it[0],col);
        }
        set<int> st;
        for(auto it : stones){
            int up = du.findParent(it[0]);
            st.insert(up);
        }
        return (stones.size()-st.size());

        
    }
};
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