<p>You are given an array <code>points</code> representing integer coordinates of some points on a 2D-plane, where <code>points[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>.</p>

<p>The cost of connecting two points <code>[x<sub>i</sub>, y<sub>i</sub>]</code> and <code>[x<sub>j</sub>, y<sub>j</sub>]</code> is the <strong>manhattan distance</strong> between them: <code>|x<sub>i</sub> - x<sub>j</sub>| + |y<sub>i</sub> - y<sub>j</sub>|</code>, where <code>|val|</code> denotes the absolute value of <code>val</code>.</p>

<p>Return <em>the minimum cost to make all points connected.</em> All points are connected if there is <strong>exactly one</strong> simple path between any two points.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/26/d.png" style="width: 214px; height: 268px;" />
<pre>
<strong>Input:</strong> points = [[0,0],[2,2],[3,10],[5,2],[7,0]]
<strong>Output:</strong> 20
<strong>Explanation:</strong> 
<img alt="" src="https://assets.leetcode.com/uploads/2020/08/26/c.png" style="width: 214px; height: 268px;" />
We can connect the points as shown above to get the minimum cost of 20.
Notice that there is a unique path between every pair of points.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> points = [[3,12],[-2,5],[-4,1]]
<strong>Output:</strong> 18
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= points.length &lt;= 1000</code></li>
	<li><code>-10<sup>6</sup> &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= 10<sup>6</sup></code></li>
	<li>All pairs <code>(x<sub>i</sub>, y<sub>i</sub>)</code> are distinct.</li>
</ul>

class DSU{
    int n;
    vector<int> parent;
    vector<int> size;
public:
    DSU(int _n){
        n=_n;
        parent.resize(n);
        for(int i=0;i<n;i++)
            parent[i]=i;
        size.resize(n,1);
    }
    
    int find(int u){
        if(u==parent[u])   return u;
        return parent[u]=find(parent[u]);
    }
    
    void Union(int a,int b){
        int parentA=find(a);
        int parentB=find(b);
        if(parentA==parentB)    return;
        if(size[parentA]>size[parentB]){
            parent[parentB]=parentA;
            size[parentA]+=size[parentB];
        }else{
            parent[parentA]=parentB;
            size[parentB]+=size[parentA];
        }
    }
    
};

class Solution {
public:
    int getDistance(vector<int> &a,vector<int>&b){
        return abs(a[0]-b[0])+abs(a[1]-b[1]);
    }
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n=points.size();
        vector<vector<int> > adj(n);
        vector<pair<int,pair<int,int> > > edges; 
        int w;
        int minCost=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(i==j)    continue;
                adj[i].push_back(j);
                if(i<j){
                    w=getDistance(points[i],points[j]);
                    edges.push_back({w,{i,j}});
                }
            }
        }
        sort(edges.begin(),edges.end());
        int maxEdges=n-1;
        int u,v;
        DSU dsu(n);
        for(int i=0;i<edges.size();i++){
            w=edges[i].first;
            u=edges[i].second.first;
            v=edges[i].second.second;
            if(dsu.find(u)!=dsu.find(v)){
                dsu.Union(u,v);
                maxEdges--;
                minCost+=w;
            }
            if(maxEdges==0) break;
        }
        return minCost;        
    }
};
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