<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> par;
    vector<int> sz;
    int find_set(int a){
        if(par[a]==a)return a;
        else return par[a]=find_set(par[a]);
    }

    void merge_union(int a,int b){
        a=find_set(a);
        b=find_set(b);
        if(a==b)return;
        if(sz[a]>sz[b])swap(a,b);
        sz[b]+=sz[a];
        par[a]=b;
    }

    vector<vector<int>> edges;
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n=points.size();
        edges.assign(n*n,vector<int>(3));
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                edges[i*n+j]={abs(points[i][0]-points[j][0])+abs(points[i][1]-points[j][1]),i,j};
            }
        }
        par.assign(n,0);
        for(int i=0;i<n;i++)par[i]=i;
        sz.assign(n,1);
        sort(edges.begin(),edges.end());
        int ans=0;
        for(auto edge:edges){
            int a=edge[1];
            int b=edge[2];
            a=find_set(a);
            b=find_set(b);
            if(a==b)continue;
            ans+=edge[0];
            merge_union(a,b);

        }
        return ans;


    }
};
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