<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>

//Visit again!

class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        int n = stones.size();
        unordered_map<int, vector<pair<int, int>>> rows, cols;
        int k = 0;
        for(auto st : stones){
            rows[st[0]].push_back({k, st[1]});
            cols[st[1]].push_back({k, st[0]});
            k++;
        }
        vector<int> visit(n, 0);
        queue<vector<int>> q;
        int ans = 0;
        for(int i=0; i<n; i++){
            if(!visit[i]){
                int res = 0;
                q.push({i, stones[i][0], stones[i][1]});
                while(!q.empty()){
                    vector<int> temp = q.front();
                    q.pop();
                    int x = temp[1], y = temp[2], idx = temp[0];
                    if(visit[idx]) continue;
                    visit[idx] = 1;
                    res++;
                    for(auto c : rows[x]){
                        if(!visit[c.first]) q.push({c.first, x, c.second});
                    }
                    for(auto r : cols[y]){
                        if(!visit[r.first]) q.push({r.first, r.second, y});
                    }
                }
                ans += res - 1;
            }
        }
        return ans;
    }
};

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