<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>

#include <vector>
#include <string>
#include <unordered_map>
#include <unordered_set>
using namespace std;

class Solution {
private:
    class Node {
        public:
            Node* parent;
            Node() {
                this->parent = nullptr;
            }
    };
    Node* find_root(Node* node) {
        Node* root = node;
        while (root -> parent != nullptr) {
            root = root->parent;
        } 
        return root;
    }
    void union_node(Node* node1, Node* node2) {
        Node* parent1 = find_root(node1);
        Node* parent2 = find_root(node2);
        if (parent1 == parent2) {
            return;
        }
        parent1->parent = parent2;
    }
    const string build_key(const int y, const int x) {
        return to_string(y) + " " + to_string(x);
    }
public:
    int removeStones(vector<vector<int>>& stones) {
        unordered_map<string, Node> m;
        unordered_map<int, string> key_same_row;
        unordered_map<int, string> key_same_col;
        for (const vector<int>& point : stones) {
            const string key = build_key(point[0], point[1]);
            m[key] = Node{};
            if (key_same_row.find(point[0]) != key_same_row.end()) {
                union_node(&m[key], &m[key_same_row[point[0]]]);
            } else {
                key_same_row[point[0]] = key;
            }
            if (key_same_col.find(point[1]) != key_same_col.end()) {
                union_node(&m[key], &m[key_same_col[point[1]]]);
            } else {
                key_same_col[point[1]] = key;
            }
        }
        unordered_set<Node*> set;
        for (const vector<int>& point : stones) {
            const string key = build_key(point[0], point[1]);
            set.insert(find_root(&m[key]));
        }
        return stones.size() - set.size();
    }
};
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