<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 {

private:
    int dfs(std::pair<int,int> current, std::unordered_map<int,std::vector<int>>& rows,
    std::unordered_map<int,std::vector<int>>& cols,  std::set<std::pair<int,int>>& visited,bool isRoot ){

        int r = current.first;
        int c = current.second;

        int res{0};

        if(visited.contains(current)) return 0;
        visited.insert(current);

        auto& rowReachable = rows[r];
        auto& colReachable = cols[c];

        for(auto el: rowReachable){
            if(el != c){
                res += dfs({r,el},rows,cols,visited,0);
            }
        }
        for(auto el: colReachable){
            if (el != r){
                res += dfs({el,c},rows,cols,visited,0);
            }
        }


        if(!isRoot) res++;
        return res;
    }

public:
    int removeStones(vector<vector<int>>& stones) {
        std::unordered_map<int,std::vector<int>> rows;
        std::unordered_map<int, std::vector<int>> cols;

        int res{0};

        for(int i = 0; i < stones.size(); i++){
            auto& point = stones[i];
            
            if(!rows.contains(point[0])){
                rows.emplace(point[0], std::vector<int>());
            }
            if(!cols.contains(point[1])){
                cols.emplace(point[1],std::vector<int>());
            }

            rows[point[0]].push_back(point[1]);
            cols[point[1]].push_back(point[0]);
        }

        std::set<std::pair<int,int>> visited;

        for(int i = 0 ; i < stones.size(); i++){
            std::pair current = {stones[i][0],stones[i][1]};

            if(!visited.contains(current)){
                res += dfs(current,rows,cols,visited,1);          
            }
        }

        return res;
    }
};
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