<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 {
public:
    int removeStones(vector<vector<int>>& stones) {
        size_t grp_cnt = 0;
        size_t n = stones.size();
        std::vector<bool> viz(n, false);
        std::unordered_map<int, std::pair<bool, std::set<size_t>>> rows;
        std::unordered_map<int, std::pair<bool, std::set<size_t>>> cols;
        std::queue<size_t> st_q;

        for(size_t i = 0; i < n; ++i) {
            if(auto r = rows.find(stones[i][0]); r != rows.end()) {
                r->second.second.insert(i);
            } else {
                rows.insert({stones[i][0], {false, std::set<size_t>()}}).first->second.second.insert(i);
            }
            if(auto c = cols.find(stones[i][1]); c != cols.end()) {
                c->second.second.insert(i);
            } else {
                cols.insert({stones[i][1], {false, std::set<size_t>()}}).first->second.second.insert(i);
            }
        }

        viz[0] = true;

        size_t i = 0;
        while(true) {
            ++grp_cnt;
            st_q.push(i);

            while(!st_q.empty()) {
                size_t cur_stone_id = st_q.front();
                st_q.pop();

                auto& stones_by_row = rows[stones[cur_stone_id][0]];
                if(!stones_by_row.first) {
                    stones_by_row.first = true;
                    for(size_t s_id: stones_by_row.second) {
                        if(viz[s_id]) continue;
                        st_q.push(s_id);
                        viz[s_id] = true;
                    }
                }

                auto& stones_by_col = cols[stones[cur_stone_id][1]];
                if(!stones_by_col.first) {
                    stones_by_col.first = true;
                    for(size_t s_id: stones_by_col.second) {
                        if(viz[s_id]) continue;
                        st_q.push(s_id);
                        viz[s_id] = true;
                    }
                }
            }

            do {
                ++i;
            } while( i < n && viz[i]);
            if(i == n) break;

        }

        return n - grp_cnt;
    }
};
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