<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 :
    vector<int>parent;
    vector<int>rank;

    DisJointSet(int val){
        parent.resize(val);
        rank.resize(val);

        for(int i = 0 ; i<val ; i++){
            parent[i] = i;
            rank[i] = 0;
        }
    }

    int findPar(int u){

        if(parent[u] == u){
            return u;
        }

        return parent[u] = findPar(parent[u]);
    }

    void unionSet(int u , int v){

        int ult_u = findPar(u);
        int ult_v = findPar(v);

        if(ult_u == ult_v){
            return;
        }

        if(rank[ult_u] > rank[ult_v]){
            parent[ult_v] = ult_u; 
        }
        else if(rank[ult_u] < rank[ult_v]){
            parent[ult_u] = ult_v; 
        }
        else{

            parent[ult_v] = ult_u;
            rank[ult_u]++;

        }

    }

};

class Solution {
public:
    int minCostConnectPoints(vector<vector<int>>& points) {

        //adjlist :
        int n = points.size();
        vector<pair<int,pair<int,int>>>adj;

        for(int i = 0 ; i<n ; i++){
            for(int j = 0 ; j<n ; j++){
                if(j!=i){
                    int wt = abs(points[i][0] - points[j][0]) + 
                    abs(points[i][1] - points[j][1]);
                    adj.push_back({wt,{i,j}});
                }
            }
        }

        sort(adj.begin(),adj.end());
        int num = adj.size();

        DisJointSet obj(num);
        int sum = 0;

        for(int i = 0 ; i< num ; i++){

            int u = adj[i].second.first;
            int v = adj[i].second.second;
            int wt = adj[i].first;

            if(obj.findPar(u) != obj.findPar(v)){
                sum+=wt;
                obj.unionSet(u,v);
            }

        }

        return sum;

        
        
    }
};
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