<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:
    int findManhatt(vector<int>point1, vector<int>point2)
    {
        int a = abs(point1[0]-point2[0]);
        int b = abs(point1[1]-point2[1]);
        return a+b;
    }
    int minCostConnectPoints(vector<vector<int>>& points) 
    {
        priority_queue<vector<int>,vector<vector<int>>,greater<vector<int>>>pq;
        int n = points.size();
        vector<int>dst(n,INT_MAX);
        dst[0]=0;
        set<int>unvisited;
        for(int i=1;i<n;i++)
        {
            unvisited.insert(i);
        }
        pq.push({0,0});
        while(pq.empty()==false)
        {
            int d = pq.top()[0];
            int idx = pq.top()[1];
            unvisited.erase(idx);
            pq.pop();
            for(auto itr = unvisited.begin();itr!=unvisited.end();itr++)
            {
                vector<int>temp = points[*itr];
                int dx = findManhatt(temp,points[idx]);
                if(dx<dst[*itr])
                {
                    dst[*itr]=dx;
                    pq.push({dx,*itr});
                }
            }
        }
        int sum=0;
        for(int i=0;i<n;i++)
        {
            //cout<<dst[i]<<endl;
            sum+=dst[i];
        }
        return sum;    
    }
};
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