<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 Solution {
public:
    vector<int> parent;
    vector<int> rank;
    int find(int u){
        if(u==parent[u]) return u;
        return parent[u]=find(parent[u]);
    }

    void Union(int u, int v){
        int uP = find(u);
        int vP = find(v);

        if(uP==vP) return;

        if(rank[uP]>rank[vP]){
            parent[vP]= uP;
        } else if(rank[vP]>rank[uP]){
            parent[uP]=vP;
        } else {
            parent[vP]=uP;
            rank[uP]++;
        }
    }

    int Kruskal(vector<vector<int>> adj){
        int sum=0;
        int V= adj.size();
        for(auto & vec: adj){
            int u = vec[0];
            int v = vec[1];
            int wt = vec[2];

            if(find(u)!=find(v)){
                Union(u,v);
                sum=sum+wt;
            }
        }
        return sum;
    }

    typedef pair<int,int> P;
    int minCostConnectPoints(vector<vector<int>>& points) {
        int V=points.size();
        rank.resize(V, 0);
        parent.resize(V);
        for(int i=0;i<V;i++){
            parent[i]=i;
        }

        vector<vector<int>> adj;
        for(int u=0;u<V;u++){
            for(int v=u+1;v<V;v++){
                int x1 = points[u][0];
                int y1= points[u][1];
                int x2 = points[v][0];
                int y2= points[v][1];
                int dist = abs(x1-x2)+abs(y1-y2);
                adj.push_back({u,v,dist});
            }
        }
        auto comporator=[&](auto & vec1, auto& vec2){
            return vec1[2]<vec2[2];
        };
        sort(begin(adj), end(adj), comporator);
        return Kruskal(adj);
    }
};
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