<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{
private:
    vector<int> parent;
    vector<int> size;

public:
    // Constructor
    DisjointSet(int n){
        this->parent.resize(n);
        // initially, every node is a parent of itself
        for(int i=0; i<n; i++)
            this->parent[i] = i;

        // initially, size of each component
        // with ultimate-parent i is 1
        this->size.resize(n, 1);
    }

    int findUltimateParent(int curr){
        if(parent[curr] == curr)
            return curr;

        parent[curr] = findUltimateParent(parent[curr]);

        return parent[curr];
    }

    // returns false if 'a' and 'b' are in the same component
    bool unionBySize(int a, int b){
        int ultPar_a = findUltimateParent(a);
        int ultPar_b = findUltimateParent(b);

        if(ultPar_a == ultPar_b)
            return false;

        if(size[ultPar_a] < size[ultPar_b])
        {
            parent[ultPar_a] = ultPar_b;
            size[ultPar_b] += size[ultPar_a];
        }
        else // size[ultPar_a] >= size[ultPar_b]
        {
            parent[ultPar_b] = ultPar_a;
            size[ultPar_a] += size[ultPar_b];
        }

        return true;
    }

    // returns true if 'a' and 'b' are in same component
    bool areConnected(int a, int b){
        return findUltimateParent(a) == findUltimateParent(b);
    }

    // num of components is given by
    // num of distinct ultimate-parents
    int componentCount(){
        int n = parent.size();
        unordered_set<int> ultPars;

        for(int i=0; i<n; i++)
            ultPars.insert(findUltimateParent(i));

        return ultPars.size();
    }
};

class Solution {
public:
    // T.C.=O(n + (maxX + maxY)), S.C.=O(maxX + maxY)
    // maxX, maxY: max val of stone[0] and stone[1] respectively
    int removeStones(vector<vector<int>>& stones) {
        int maxX = 0, maxY = 0;
        for(vector<int>& stone: stones)
        {
            maxX = max(maxX, stone[0]);
            maxY = max(maxY, stone[1]);
        }

        DisjointSet ds(maxX + 1 + maxY + 1);
        vector<bool> isPresent(maxX + 1 + maxY + 1, false);
        for(vector<int>& stone: stones)
        {
            int a = stone[0];
            // offset the y coordinate by (maxX + 1)
            int b = maxX + 1 + stone[1];

            ds.unionBySize(a, b);

            isPresent[a] = true;
            isPresent[b] = true;
        }

        int absentNodeCount = count(isPresent.begin(), isPresent.end(), false);
        return stones.size() - (ds.componentCount() - absentNodeCount);
    }
};

/*
# using rows and cols as nodes of a graph
# num of nodes in graph must be ((maxX + 1) + (maxY + 1))
# row coordinate remains the same for a node but, 
  col coordinate is offset by (maxX + 1)
# to get the actual component count: 
  remove the nodes which don't have any stone lying on them
*/
30101