<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> parent;
    void make(){
        int n = parent.size();
        for(int i=1;i<n;i++)parent[i] = i;
    }

    int find(int u){
        if(u == parent[u])return u;

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

    void Union(int u,int v){
        parent[u] = v;
    }
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n = points.size();
        vector<vector<pair<int ,int>>> graph(n+1);
        parent.resize(n+1);
        make();
        
        int maxDist = INT_MAX;
        vector<int> pointsMaxDis;
        map<pair<int,int>, int> pointMap;
        int k = 1;

        priority_queue <pair<int,pair<int,int>>, vector<pair<int,pair<int,int>>>, greater<pair<int,pair<int,int>>>> pq;
        

        for(int i=0;i<n;i++){
            int u = points[i][0] , v = points[i][1];
            pointMap[{u,v}] = k++;
        }

        for(int i=0;i<n;i++){
            int u = points[i][0], v = points[i][1];
            for(int j = i+1;j<n;j++){
                int a = points[j][0] , b =points[j][1];
                int dist = abs(u-a) +abs(v-b);
                graph[pointMap[{u,v}]].push_back({dist,pointMap[{a,b}]});
                pq.push({dist,{pointMap[{u,v}],pointMap[{a,b}]}});
            }
        }
        int cost = 0 ;
        while(!pq.empty()){
            pair<int,pair<int,int>> curr = pq.top();
            pq.pop();
            int u = curr.second.first;
            int v = curr.second.second;

            int parent_u = find(u);
            int parent_v = find(v);
            if(parent_u == parent_v)continue;
            else{
                Union(parent_u,parent_v);
                cost += curr.first;
            }
        }
        return cost ;        
    }
};
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