<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 {
    int findParent(vector<int>& parent, int ind){
        if(ind == parent[ind]){
            return ind;
        }
        return parent[ind]=findParent(parent, parent[ind]);
    }
    void unionSet(vector<int>& parent, vector<int>& size, int ind1, int ind2){
        int p1=findParent(parent, ind1);
        int p2=findParent(parent, ind2);
        if(p1==p2){
            return;
        }else if(size[p1]>size[p2]){
            swap(p1, p2);
        }
        size[p2]+=size[p1];
        parent[p1]=p2;
    }
public:
    int removeStones(vector<vector<int>>& stones) {
        int n=stones.size();
        vector<int> parent(n, 0);
        vector<int> size(n, 1);
        for(int i=0;i<n;i++){
            parent[i]=i;
        }
        for(int i=0;i<n;i++){
            for(int j=1;j<n;j++){
                if(stones[i][0]==stones[j][0] || stones[i][1]==stones[j][1]){
                    unionSet(parent, size, i, j);
                }
            }
        }
        int c=0;
        for(int i=0;i<n;i++){
            c+=(parent[i]==i);
        }
        return n-c;
    }
};
18238