<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 DSU {
public:
    vector<int> parent, size;

    // constructor to make the individual nodes
    DSU(int n) {
        for (int i = 0; i <= n; i++) {
            // make them their own parent
            parent.push_back(i);
            // make their size to 1
            size.push_back(1);
        }
    }

    int find(int n) {
        // if parent is same as the node then return the parent/node
        if (parent[n] == n)
            return parent[n];
        // else store the ultimate parent to that node
        // this (parent[n] = ) is path compression
        return parent[n] = find(parent[n]);
    }

    void dsuBySize(int a, int b) {
        // storing the value of their parent to themselves
        a = find(a), b = find(b);
        if (a != b) {
            // making sure that a always have the bigger value
            if (b > a)
                swap(a, b);
            // joining the smaller parent to bigger parent
            parent[b] = a;
            // increase the bigger nodes size
            size[a] += size[b];
        }
    }
};

class Solution {
public:
    int removeStones(vector<vector<int>>& stones) {
        int maxRow = 0, maxCol = 0;
        int n = stones.size();
        for (auto it : stones) {
            maxRow = max(maxRow, it[0]);
            maxCol = max(maxCol, it[1]);
        }
        // initialize dsu
        DSU ds(maxRow + maxCol + 1);
        // storing the node counts
        int c = maxRow + maxCol + 1;

        for (auto it : stones) {
            int nRow = it[0];
            int nCol = it[1] + maxRow + 1;
            ds.dsuBySize(nRow, nCol);
        }

        for (int i = 0; i <= c; i++) {
            // if the parent is same as them and the node have child node in
            // them then decrease the n
            if (ds.find(i) == i && ds.size[i] > 1)
                n--;
        }
        return n;
    }
};
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