<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>

#define PI pair<int,int>

void print_pairList(string msg, vector<PI> &pList) {
    cout<<endl<<msg;
    for(auto& p : pList) {
        cout<<"{"<<p.first<<","<<p.second<<"} ";
    }
}

class Solution {
public:
    void init_graph(vector<vector<int>>& stones, map<int, vector<PI>>& rowList, map<int, vector<PI>>& colList) {
        for(auto& stone : stones) {
            PI stonePair = {stone[0], stone[1]};
            colList[stone[0]].push_back(stonePair);
            rowList[stone[1]].push_back(stonePair);
        }
    }

    vector<PI> getNeighbours(PI stone, map<int, vector<PI>>& rowList, map<int, vector<PI>>& colList, set<PI>& visited) {
        vector<PI> neighbours;
        for(auto neighbour : colList[stone.first]) {
            if(!visited.count(neighbour))
                neighbours.push_back(neighbour);
        }
        for(auto neighbour : rowList[stone.second]) {
            if(!visited.count(neighbour))
                neighbours.push_back(neighbour);
        }
        return neighbours;
    }

    void dfs(PI stone, map<int, vector<PI>>& rowList, map<int, vector<PI>>& colList, set<PI>& visited) {
        // cout<<"\n"<<stone.first<<","<<stone.second;
        visited.insert(stone);
        vector<PI> neighbours = getNeighbours(stone, rowList, colList, visited);
        for(auto neighbour : neighbours) {
            dfs(neighbour, rowList, colList, visited);
        }
    }

    int removeStones(vector<vector<int>>& stones) {
        map<int, vector<PI>> rowList;
        map<int, vector<PI>> colList;
        init_graph(stones, rowList, colList);
        // print_pairList("col0", colList[0]);
        // print_pairList("col1", colList[1]);
        // print_pairList("col2", colList[2]);
        // print_pairList("row0", rowList[0]);
        // print_pairList("row1", rowList[1]);
        // print_pairList("row2", rowList[2]);
        // cout<<"\ncol "<<colList.size();
        // cout<<"\nrow "<<rowList.size();
        set<PI> visited;
        int numConnectedComponents = 0;
        for(auto& stone : stones) {
            PI stonePair = {stone[0], stone[1]};
            if(!visited.count(stonePair)) {
                // cout<<"\n-------- new DFS ----------";
                dfs(stonePair, rowList, colList, visited);
                numConnectedComponents++;
            }
        }
        // cout<<"\nvisited: "<<visited.size();
        // cout<<"\nnumComp: "<<numConnectedComponents;
        return stones.size() - numConnectedComponents;
    }
};
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