<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>

struct Edge{
   int src;
   int des;
    int wt;
};
class Solution {
public:
    static bool cmp(Edge e1,Edge e2){
        return e1.wt<e2.wt;
    }
    int find(vector<int>&parent,int x){
    // O(log*n)
    return parent[x]= (parent[x]==x)? x :(find(parent,parent[x]));
}
void Union(vector<int>&rank,vector<int>&parent,int a,int b){
    // O(log*n)
     a = find(parent,a);
     b = find(parent,b);
     if(a==b) return;
    if(rank[a]<rank[b]){
        rank[b]++;
        parent[a] = b;
    }else{
        rank[a]++;
        parent[b] = a;
    }
}
    int manhatten(vector<int>&a,vector<int>&b){
        return abs(a[0]-b[0]) + abs(a[1]-b[1]);
    }
    int kruskal(vector<Edge>&input,int v){
        sort(input.begin(),input.end(),cmp);
        vector<int>parent(v);
        vector<int>rank(v,0);
        for(int i =0;i<v;i++){
            parent[i] = i;
        }
        int edgecount  =0;
        int i =0;
        int ans =0;
        while(i<input.size() and edgecount<v-1){
            Edge curr = input[i];
            int srcp = find(parent,curr.src);
            int desp = find(parent,curr.des);
            if(srcp != desp){
                ans += curr.wt;
                Union(rank,parent,srcp,desp);
                edgecount++;
            }
            i++;
        }
        return ans;
    }
    int minCostConnectPoints(vector<vector<int>>& points) {
        int v = points.size();
        int e = v*(v-1)/2;
        vector<Edge>input(e);
        int k = 0;
        for(int i =0;i<points.size();i++){
            for(int j = i+1;j<points.size();j++){
                int w = manhatten(points[i],points[j]);
                input[k].src = i;
                input[k].des = j;
                input[k].wt = w;
                k++;
            }
        }
        return kruskal(input,v);
    }
};
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