<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>

//T.C : O(n^2 * α(n))
//S.C : O(n)

//PLAIN TEMPLATE FOR USAGE
class DisjointSet{
    vector<int> parent;
    vector<int> rank;

    public:
    DisjointSet(int n){
        rank.resize(n+1 , 0);
        parent.resize(n+1);

        for(int i=0;i<=n;i++)
            parent[i] = i;
        
    }

    int findParent(int x)
    {
        if(x == parent[x])
            return x;

        return parent[x] = findParent(parent[x]);
    }

    //UNION BY RANK CODE:
    //BASED ON IDEA THAT LOWER RANK NODE WILL BE CONNECTED TO HIGHER RANK NODE
    void Union_By_Rank( int x , int y )
    {
        int x_parent = findParent(x);
        int y_parent = findParent(y);

        if(x_parent == y_parent)
            return;

        //Parents are not same meaning the nodes are from different graphs
        if(rank[x_parent] == rank[y_parent]){
            parent[x_parent] = y_parent;
            rank[y_parent]++;
        }
        else if(rank[x_parent] < rank[y_parent]){
            parent[x_parent] = y_parent;
        }
        else if(rank[x_parent] > rank[y_parent]){
            parent[y_parent] = x_parent;
        }
    }

};

class Solution {
public:    
    int removeStones(vector<vector<int>>& stones) {
        int n = stones.size();
        DisjointSet S(n);
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                if(stones[i][0] == stones[j][0] || stones[i][1] == stones[j][1])
                    S.Union_By_Rank(i , j);
            }
        }

        int grpCtr = 0;
        for(int i=0;i<n;i++){
            if(S.findParent(i) == i)
                grpCtr++;
        }

        return (n - grpCtr);
    }
};

18056