<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> parent,size;
  DisjointSet(int n) {
    parent.resize(n+1);
    size.resize(n+1);

    for(int i=0;i<=n;i++) {
      parent[i] = i;
      size[i] = 1;
    }
  }

  int findUPar(int node) {
    if(parent[node] == node) return node;

    return parent[node] = findUPar(parent[node]);
  }
  
  void unionBySize(int u, int v) {
    int ulp_u = findUPar(u);
    int ulp_v = findUPar(v);

    if(ulp_u == ulp_v) return;
    if(size[ulp_u] < size[ulp_v]) {
      parent[ulp_u] = ulp_v;
      size[ulp_v] += size[ulp_u];
    } 
    else {
      parent[ulp_v] = ulp_u;
      size[ulp_u] += size[ulp_v];
    }
  }
};

class Solution {
public:
  int removeStones(vector<vector<int>>& stones) {
    int n = stones.size();
    int maxRow=0;
    int maxCol=0;

    for(auto it:stones) {
      maxRow = max(maxRow, it[0]);
      maxCol = max(maxCol, it[1]);
    }

    DisjointSet ds(maxRow + maxCol + 1);

    unordered_map<int,int> stoneNodes;

    for(auto it:stones) {
      int nodeRow = it[0];
      int nodeCol = it[1] + maxRow + 1;

      ds.unionBySize(nodeRow, nodeCol);
      stoneNodes[nodeRow] = 1;
      stoneNodes[nodeCol] = 1;
    }

    int cnt=0;
    for(auto it: stoneNodes) {
      if(ds.findUPar(it.first) == it.first) cnt++;
    }

    return n-cnt;
  }
};
24261