<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>

static const int fast = []{
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);

    return 0;
}();
class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        int n = stones.size();
        int ans = 0;
        vector<int> vis(n, 0);  // Use a 1D visited array based on the stone indices

        for (int i = 0; i < n; ++i) {
            if (!vis[i]) {  // If the stone at index i is not visited
                ans++;  // Increment the count for a new connected component
                queue<int> q;
                q.push(i);
                vis[i] = 1;

                while (!q.empty()) {
                    int idx = q.front();
                    q.pop();

                    // Iterate over all stones to find connected ones
                    for (int j = 0; j < n; ++j) {
                        if (!vis[j] && (stones[j][0] == stones[idx][0] || stones[j][1] == stones[idx][1])) {
                            q.push(j);
                            vis[j] = 1;
                        }
                    }
                }
            }
        }

        // Return the number of stones minus the number of connected components
        return n - ans;
    }
};
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