<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:
    void dfs(const map<pair<int, int>, vector<pair<int, int>>>& adj, const pair<int, int>& node, map<pair<int, int>, bool>& vis, int& cnt) {
        cnt++;
        vis[node] = true;
        auto it = adj.find(node);
        if (it != adj.end()) {
            for (const auto& neighbor : it->second) {
                if (!vis[neighbor]) {
                    dfs(adj, neighbor, vis, cnt);
                }
            }
        }
    }

    int removeStones(vector<vector<int>>& stones) {
        int n = stones.size();
        map<pair<int, int>, vector<pair<int, int>>> adj;
        map<int, vector<int>> rows;
        map<int, vector<int>> columns;

        // Build adjacency list
        for (const auto& stone : stones) {
            int row = stone[0];
            int col = stone[1];
            rows[row].push_back(col);
            columns[col].push_back(row);
        }

        // Add edges for rows
        for (const auto& [row, cols] : rows) {
            for (int i = 0; i < cols.size() - 1; ++i) {
                adj[{row, cols[i]}].push_back({row, cols[i + 1]});
                adj[{row, cols[i + 1]}].push_back({row, cols[i]});
            }
        }

        // Add edges for columns
        for (const auto& [col, rows] : columns) {
            for (int i = 0; i < rows.size() - 1; ++i) {
                adj[{rows[i], col}].push_back({rows[i + 1], col});
                adj[{rows[i + 1], col}].push_back({rows[i], col});
            }
        }

        // DFS to find connected components
        map<pair<int, int>, bool> vis;
        int ans = 0;

        for (const auto& stone : stones) {
            pair<int, int> node = {stone[0], stone[1]};
            if (!vis[node]) {
                int cnt = 0;
                dfs(adj, node, vis, cnt);
                ans += cnt - 1;
            }
        }

        return ans;
    }
};

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