<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 DisjointSet {
    private:
    vector<int> parent;
    vector<int> rank;

    public:
    DisjointSet(int n) {
        parent.resize(n);
        rank.resize(n, 0);
        for(int i = 0; i < n; i++) parent[i] = i;
    }

    int find(int x) {
        if (parent[x] != x) {
            parent[x] = find(parent[x]);
        }
        return parent[x];
    }

    void unionSets(int x, int y) {
        int rootX = find(x);
        int rootY = find(y);

        if (rootX != rootY) {
            if (rank[rootX] > rank[rootY]) {
                parent[rootY] = rootX;
            } else if (rank[rootX] < rank[rootY]) {
                parent[rootX] = rootY;
            } else {
                parent[rootY] = rootX;
                rank[rootX]++;
            }
        }
    }
};

class Solution {
public:
    pair<int, int> findMax(vector<vector<int>>& stones, int n) {
        int maxRow = INT_MIN, maxCol = INT_MIN;

        for(int i = 0; i < n; i++) {
            if (stones[i][0] > maxRow) maxRow = stones[i][0];
            if (stones[i][1] > maxCol) maxCol = stones[i][1];
        }

        return { maxRow, maxCol };
    }

    int removeStones(vector<vector<int>>& stones) {
        int e = stones.size(), n, x, y, count = 0;
        pair<int, int> max = findMax(stones, e);
        n = max.first + max.second + 2;
        DisjointSet ds(n);
        vector<bool> hashmap(n, false);

        for(int i = 0; i < e; i++) {
            x = stones[i][0];
            y = stones[i][1] + max.first + 1;

            if (ds.find(x) != ds.find(y)) ds.unionSets(x, y);
        }

        for(int i = 0; i < n; i++) {
            if (ds.find(i) != i && !hashmap[ds.find(i)]) {
                hashmap[ds.find(i)] = true;
                count++;
            }
        }
        cout<<count;

        return e - count;
    }
};
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