<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>


template <class T>
class DisjointSetDS {

    unordered_map<T,T> par;
    unordered_map<T, pair<int,int>> representative_to_set_info;

public:

    void make_set(T x) {
        par[x] = x;
        representative_to_set_info[x] = {1,0}; // size, height
    }

    T find(T x) {
        if(x == par[x]) {
            return x;
        }
        return par[x] = find(par[x]);
    }

    void unionSet(T u, T v) {
        T ru = find(u);
        T rv = find(v);
        if(ru == rv) {
            return;
        }

        int hu = representative_to_set_info[ru].second;
        int hv = representative_to_set_info[rv].second;

        if(hu > hv) {
            par[rv] = ru;
            representative_to_set_info[ru].first += representative_to_set_info[rv].first;
            representative_to_set_info.erase(rv);
        } else {
            par[ru] = rv;
            representative_to_set_info[rv].first += representative_to_set_info[ru].first;
            representative_to_set_info.erase(ru);
            if(hu == hv) {
                representative_to_set_info[rv].second++;
            }
        }
    }

    int components() {
        return representative_to_set_info.size();
    }

};


class Solution {
public:

    int removeStones(vector<vector<int>>& stones) {
        // DSU.
        int n = stones.size();

        DisjointSetDS<int> dsu;

        map<int, vector<int>> row, col;

        for(int i=0; i<n; i++) {
            int r = stones[i][0], c = stones[i][1];
            row[r].push_back(i);
            col[c].push_back(i);
            dsu.make_set(i);
        }

        for(auto it:row) {
            vector<int> vc = it.second;
            int f = vc[0];
            for(int j=1; j<vc.size(); j++) {
                dsu.unionSet(f, vc[j]);
            }
        }
        for(auto it:col) {
            vector<int> vc = it.second;
            int f = vc[0];
            for(int j=1; j<vc.size(); j++) {
                dsu.unionSet(f, vc[j]);
            }
        }

        int ans = dsu.components();
        return (n - ans);
    }
};
31013