<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>

bool compare(vector<int>&a,vector<int>&b){
        return a[2]<b[2];
    }
class Solution {
    
    int findParent(vector<int>&parent,int vertex){
        if(parent[vertex]==vertex){
            return vertex;
        }
        return parent[vertex]=findParent(parent,parent[vertex]);
    }
    void Union(vector<int>&parent,vector<int>&rank,int u,int v){
        u=findParent(parent,u);
        v=findParent(parent,v);
        if(rank[u] >= rank [v]){
            parent[v]=u;
            rank[u]++;
        }
        else if(rank[u] < rank[v]){
            parent[u]=v;
            rank[v]++;
        }
    }
    int kruskal(vector<vector<int> >&edgeList,int n){
        vector<int>parent(n);
        vector<int>rank(n,0);

        for(int i=0;i<n;i++){
            parent[i]=i;
        }
        sort(edgeList.begin(),edgeList.end(),compare);
        int ans=0;
        int edgesCount=0;
        for(int i=0;(i<edgeList.size()) && (edgesCount < (n-1)) ;i++){
            int u=findParent(parent,edgeList[i][0]);
            int v=findParent(parent,edgeList[i][1]);
            int wt=edgeList[i][2];
            if(u!=v){
                edgesCount++;
                ans+=wt;
                Union(parent,rank,u,v);
            }
        }
        return ans;
    }
public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n=points.size();
        vector<vector<int> > edgeList;


        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                int u=i;
                int v=j;
                int wt=abs(points[i][0]-points[j][0])+abs(points[i][1]-points[j][1]);
                vector<int>temp;
                temp.push_back(u);
                temp.push_back(v);
                temp.push_back(wt);
                edgeList.push_back(temp);
            }
        }

        return kruskal(edgeList,n);
    }
};
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