<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,size;
    DisjointSet(int n){
        rank.resize(n+1,0);
        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(node==parent[node])
            return node;
            
        return parent[node]=findUPar(parent[node]);
    }
    
    void unionByRank(int u, int v){
        int ulp_u=findUPar(u);
        int ulp_v=findUPar(v);
        
        if(ulp_u==ulp_v) return;
        
        if(rank[ulp_u]<rank[ulp_v]){
            parent[ulp_u]=ulp_v;
        }
        else if(rank[ulp_u]>rank[ulp_v]){
            parent[ulp_v]=ulp_u;
        }
        else{
            parent[ulp_v]=ulp_u;
            rank[ulp_u]++;
        }
    }
    
    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) {
        //===================Connect ROW-COL==========================
        
        // int maxRow=0;
        // int maxCol=0;
        // int n=stones.size();
        // for(auto it: stones){
        //     maxRow=max(maxRow,it[0]);
        //     maxCol=max(maxCol,it[1]);
        // }

        // DisjointSet ds(maxRow+maxCol+1);

        // for(auto it: stones){
        //     int row=it[0];
        //     int col=it[1]+maxRow+1;

        //     ds.unionBySize(row,col);

        // }

        // int cnt=0;
        // for(int i=0;i<maxRow+maxCol+1;i++){
        //     if(ds.parent[i]==i && ds.size[i]!=1)
        //         cnt++;
        // }

        // return n-cnt;

        //===========================Connect stones================
        int n = stones.size();
        DisjointSet ds(n);
        unordered_map<int, int> rowMap, colMap;

        // Union all stones that share the same row or column
        for (int i = 0; i < n; i++) {
            int row = stones[i][0];
            int col = stones[i][1];

            // If this row already has a stone, union this stone with that stone
            if (rowMap.count(row)) {
                ds.unionBySize(i, rowMap[row]);
            }
            // If this column already has a stone, union this stone with that stone
            if (colMap.count(col)) {
                ds.unionBySize(i, colMap[col]);
            }

            // Map the current stone to the corresponding row and column
            rowMap[row] = i;
            colMap[col] = i;
        }

        // Count the number of unique connected components (root nodes)
        int connectedComponents = 0;
        for (int i = 0; i < n; i++) {
            if (ds.parent[i] == i) {
                connectedComponents++;
            }
        }

        // Maximum stones we can remove is (total stones - number of connected components)
        return n - connectedComponents;
    }
};
20793