<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:

    class Compare{
        public:
            bool operator()(pair<pair<int,int>, int> p1, pair<pair<int,int>, int> p2){
                return p1.second>p2.second;
            }
    };

    int findparent(int p, vector<int> parents){
        if(p!=parents[p]){
            int parent = findparent(parents[p], parents);
            parents[p] = parent;
        }
        return parents[p];
    }

    int minCostConnectPoints(vector<vector<int>>& points) {
       int n = points.size();
       vector<int> groups(n,0);
       vector<int> groupsize(n,1);

       for(int i=0;i<n;i++){
            groups[i] = i;
       }

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

        for(int i=0;i<n;i++){
            auto point = points[i];
            for(int j=i+1;j<n;j++){
                pq.push({{i,j}, abs(points[i][0]-points[j][0]) + abs(points[i][1]-points[j][1])});
            }
        }
        int result = 0;

        while(pq.size()>0){
            auto top = pq.top();
            int p1 = top.first.first;
            int p2 = top.first.second;
            int dist = top.second;
            cout<<p1<<" "<<p2<<" "<<dist<<endl;
            pq.pop();
            int p1parent = findparent(p1, groups);
            int p2parent = findparent(p2, groups);

            if(p1parent!=p2parent){
                if(groupsize[p1parent]>groupsize[p2parent]){
                    groups[p2parent] = p1parent;
                    groupsize[p1parent]+=groupsize[p2parent];
                } else {
                    groups[p1parent] = p2parent;
                    groupsize[p2parent]+=groupsize[p1parent];
                }
                result += dist;
                // cout<<groupsize[p1parent]<<" "<<groupsize[p2parent]<<endl;
                if(groupsize[p1parent]==n || groupsize[p2parent]==n){
                    break;
                }
            }
        }

        return result;
    }
};
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