<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>

typedef pair<int,int> p;

class Solution {
public:

    p findpar(map<p,p>&parmap, p x){
        p val = parmap[x];
        if(val.first == x.first && val.second == x.second)
            return val;
        return parmap[x] = findpar(parmap,val);
    }

    void makeunion(map<p,p>&parmap, map<p,int>&rankmap, p x, p y){

        p upx = findpar(parmap,x);
        p upy = findpar(parmap,y);

        if(upx.first == upy.first && upx.second == upy.second)
            return;

        if(rankmap[upx] > rankmap[upy]) parmap[upy] = upx;
        else if(rankmap[upx] == rankmap[upy]){
            parmap[upy] = upx;
            rankmap[upx]++;
        }
        else parmap[upx] = upy;
    }

    int removeStones(vector<vector<int>>& stones) {
        int n = stones.size();
        if(n==1)
            return 0;

        map<p,p> parmap;
        map<p,int> rankmap;
        
        // Initializing the map
        for(int i=0;i<n;i++){
            p u;
            u.first = stones[i][0];
            u.second = stones[i][1];
            parmap[u] = u;
            rankmap[u] = 1;
        }

        for(int i=0;i<n;i++){
            p u;
            u.first = stones[i][0];
            u.second = stones[i][1];
            for(int j=i+1; j<n; j++){
                p v;
                v.first = stones[j][0];
                v.second = stones[j][1];
                if(u.first == v.first || u.second == v.second)
                    makeunion(parmap,rankmap,u,v);
            }
        }

        for(int i=0;i<n;i++) {
            p u;
            u.first = stones[i][0];
            u.second = stones[i][1];
            parmap[u] = findpar(parmap,u);
        }

        map<p,int> countmap;
        for(auto it : parmap){
            // cout<<it.first.first<<" "<<it.first.second<<" ===>";
            // cout<<it.second.first<<" "<<it.second.second<<endl;
            countmap[it.second]++;
        }

        int removed = 0;
        for(auto it : countmap){
            removed += it.second-1;
        }

        return removed;
    }
};
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