<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, 1);
        for(int i=0;i<=n;i++){
            parent[i] = i;
        }
    }
    int findUParent(int node){
        if(parent[node] == node){
            return node;
        }
        return parent[node] = findUParent(parent[node]);
    }
    void unionBySize(int u, int v){
        int u_parent = findUParent(u);
        int v_parent = findUParent(v);
        if(u_parent == v_parent) return;
        if(size[u_parent] < size[v_parent]){
            parent[u_parent] = v_parent;
            size[v_parent] += size[u_parent];
        }else{
            parent[v_parent] = u_parent;
            size[u_parent] += size[v_parent];
        }
    }
};

class Solution {
public:
//Note - this question base logic is to get the no. of component.
//       but key point is node(stones) can be a part of same component if there,
//       row/col is same.
//       so this will get little complex or will be very bruthforce if we make node as stones
//       instead will try to make each row and col as a node and then will take,
//       row and col of stone and do union of the same(since node are row and col).
//       here to do this we need to map the coordinate of col since row will take starting digits
//       ex if we have 5*4 grid, then 0-4 will be row and 5-8 will be col.
    int removeStones(vector<vector<int>>& stones) {
        int maxRow = 0;
        int maxCol = 0;
        int totalStones = 0;
        for(auto it: stones){
            totalStones++;
            maxRow = max(maxRow, it[0]);
            maxCol = max(maxCol, it[1]);
        }
        int n = maxRow + maxCol;
        DisJointSet ds(n+1);
        for(auto it: stones){
            int r = it[0];
            int c = it[1]+maxRow+1;//coordinate mapping
            ds.unionBySize(r, c);
        }
        int cnt = 0;
        for(int i=0;i<=n+1;i++){
            if(ds.parent[i] == i && ds.size[i]>1) cnt++;
        }
        return totalStones-cnt;
    }
};
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