<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 DisjointSet{
    public:
    int n;
    vector<int> parent;
    vector<int> rank;
    DisjointSet(int n){
        this->n = n;
        parent.resize(n);
        rank.resize(n);
        for(int i = 0 ; i <n ; i++){
            parent[i] = i;
            rank[i] = 0;
        }
    }
    void unionByRank(int u, int v){
        int par1 = findUpar(u);
        int par2 = findUpar(v);
        if(rank[par1] >= rank[par2]){
            parent[par2] = par1;
            rank[par1]++;
        }else{
            parent[par1] = par2;
            rank[par2]++;
        }
    }
    int findUpar(int u){
        if(parent[u] == u)return u;
        else return parent[u] = findUpar(parent[u]);
    }
};
class Solution {
public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n = points.size();
        vector<vector<int>> edges;
        edges.reserve((n*(n-1)) / 2);
        for(int i = 0 ; i < n; i++){
            for(int j = i+1 ; j < n ; j++){
                int dist = abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1]);
                edges.push_back({dist,i,j});
            }
        }
        sort(edges.begin(), edges.end());
        DisjointSet ds(n);
        int cnt = 0;
        int i = 0, sum = 0;
        while(cnt != n-1){
            int wt= edges[i][0];
            int u = edges[i][1];
            int v = edges[i][2];
            i++;
            if(ds.findUpar(u) == ds.findUpar(v))continue;
            ds.unionByRank(u, v);
            sum += wt;
            cnt++;
        }
        return sum;

    }
};
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