<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 DisjointSet{
    public:
        vector<int> parent;
        vector<int> size;

        DisjointSet(int n){
            parent.resize(n+1);
            size.resize(n+1,1);
            for(int i = 0;i<n+1;i++){
                parent[i] = i;
            }

        }

        int findUpar(int u){
            if(parent[u]==u){
                return u;
            }
            return parent[u] = findUpar(parent[u]);
        }

        void unionBySize(int u,int v){
            int ult_u = findUpar(u);
            int ult_v = findUpar(v);

            if(ult_u==ult_v){
                return;
            }
            else if(size[ult_u]>size[ult_v]){
                parent[ult_v] = ult_u;
                size[ult_u]+=size[ult_v];
            }
            else{
                parent[ult_u] = ult_v;
                size[ult_v]+=size[ult_u];
            }
        }
};

class Solution {

public:
    int removeStones(vector<vector<int>>& stones) {
         int n = stones.size();
         int maxRow = 0;
         int maxCol = 0;
         for(int i = 0;i<n;i++){
             maxRow = max(maxRow,stones[i][0]);
             maxCol = max(maxCol,stones[i][1]);
         }

         int sz = maxRow+maxCol+1;
         DisjointSet ds(sz);
         for(auto it : stones){
             int row = it[0];
             int col = it[1]+maxRow+1;
             ds.unionBySize(row,col);
         }

         int ans = 0;
         for(int i = 0;i<=maxRow+maxCol+1;i++){
            if(ds.parent[i]==i && ds.size[i] > 1){
                 ans++;
             }
         }
        
         return n-ans;
    }
};
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