<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 {
public:
    vector<int> rank, parent;

    DisjointSet(int n) {
        rank.resize(n + 1, 0);
        parent.resize(n + 1);
        for (int i = 0; i <= n; i++) {
            parent[i] = i;
        }
    }

    int findpar(int x) {
        if (x == parent[x])
            return x;
        return parent[x] = findpar(parent[x]);  // Path compression
    }

    void u_by_r(int a, int b) {
        int ua = findpar(a);
        int ub = findpar(b);
        
        if (ua != ub) {  // Only union if they are in different sets
            if (rank[ua] < rank[ub])
                parent[ua] = ub;
            else if (rank[ua] > rank[ub])
                parent[ub] = ua;
            else {
                parent[ub] = ua;
                rank[ua]++;
            }
        }
    }
};

class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        int idx = 0, jdx = 0;

        // Find maximum row and column indices
        for (int i = 0; i < stones.size(); i++) {
            idx = max(idx, stones[i][0]);
            jdx = max(jdx, stones[i][1]);
        }

        DisjointSet ds(idx + jdx + 2);  // +2 to avoid off-by-one errors
        unordered_map<int, int> M;

        for (int i = 0; i < stones.size(); i++) {
            int row = stones[i][0];
            int col = stones[i][1] + idx + 1;  // Offset columns
            ds.u_by_r(row, col);
            M[row] = 1;
            M[col] = 1;
        }

        int res = 0;

        // Count unique components (roots)
        for (auto it : M) {
            if (ds.findpar(it.first) == it.first)
                res++;
        }

        return stones.size() - res;
    }
};

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