<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:
typedef pair<int,int> ppi ; 
    int dist(vector<int>a,vector<int>b){
         int dori = abs(b[1]-a[1]) + abs(a[0]-b[0]);
        return dori ;
    }
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n = points.size() ;
        // making adj (considering each point as a node)
        unordered_map<int,vector<pair<int,int>>> adj ; 
        for(int i =0 ; i  < n ; i++){
            for(int j = i+1;j<n;j++){
                int distance = dist(points[i],points[j]);
                adj[i].push_back({j,distance});
                adj[j].push_back({i,distance});
            }
        }
        // necessary stuff 
        int sum = 0 ; 
        vector<bool>visi(n,0); // visited arr  
        priority_queue<ppi , vector<ppi> , greater<ppi>>pq ;
        pq.push({0,0}); // distance  , node ; 

        while(!pq.empty()){
            auto p = pq.top();
            pq.pop();
            int node = p.second ;
            int dist = p.first ; 
            if(visi[node]) continue ; 
            visi[node] = true ; 
            sum += dist ; 
            for(auto nei : adj[node]){
                int v = nei.first ; 
                int c = nei.second ;
                if(!visi[v]){
                    pq.push({c,v});
                }
            } 
        }
        return sum ; 

    }
};
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