<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 find(vector<int> p, int e) {
        if(p[e] == e) {
            return e;
        }
        return (p[e] = find(p, p[e]));
    }
    int mod(int n) {
        return n<0?-n:n;
    }
    int distance(vector<vector<int>> &points, int i, int j) {
        return mod(points[i][0]-points[j][0]) + mod(points[i][1]-points[j][1]);
    }

    int minCostConnectPoints(vector<vector<int>>& points) {
        vector<int> p(points.size()+1);
        vector<int> count(points.size()+1, 1);
        for(int i=0; i<points.size(); i++) {
            p[i] = i;
        }

        priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<pair<int, pair<int, int>>>> qs;
        for(int i=0; i<points.size(); i++) {
            for(int j=i+1; j<points.size(); j++) {
                qs.push(make_pair(distance(points, i, j), make_pair(i, j)));
            }
        }

        int mxs = 1;
        int sum = 0;
        while(mxs < points.size()) {
            pair<int, pair<int, int>> pr = qs.top();
            qs.pop();
            int fs = find(p, pr.second.first);
            int ss = find(p, pr.second.second);
            if(fs == ss) {
                continue;
            }
            sum += pr.first;
            if(count[fs] > count[ss]) {
                p[ss] = fs;
                count[fs] += count[ss];
                mxs = max(mxs, count[fs]);
            } else {
                p[fs] = ss;
                count[ss] += count[fs];
                mxs = max(mxs, count[ss]);
            }
        }
        return sum;
    }
};
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