<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 DisjointSet{
public:
    vector<int> root, rank;
    int cnt;
    
    DisjointSet(int size): root(size), rank(size, 1), cnt(size){
        for (int i = 0; i < size; i++){
            root[i] = i;
        }
    }
    
    int find(int x){
        if (x == root[x]){
            return x;
        }
        root[x] = find(root[x]);
        return root[x];
    }
    
    void unionSet(int x, int y){
        int rx = find(x), ry = find(y);
        if (rx != ry){
            if (rank[rx] > rank[ry]){
                root[ry] = rx;
            } else if(rank[rx] < rank[ry]){
                root[rx] = ry;
            } else {
                root[ry] = rx;
                rank[rx] += 1;
            }
            cnt--;
        }
    }
    
    bool connected(int x, int y){
        return find(x) == find(y);
    }
    
    int count(){
        return cnt;
    }
};

class Solution {
public:
    static int comp(vector<int>& a, vector<int>& b){
        return a[0] < b[0];
    }

    int minCostConnectPoints(vector<vector<int>>& points) {
        vector<vector<int>> edges;
        for (int i = 0; i < points.size(); i++){
            for (int j = i+1; j < points.size(); j++){
                if (i != j){
                    int dist = abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1]);
                    vector<int> edge = {dist, i, j};
                    edges.push_back(edge);
                }
            }
        }
                 
        sort(edges.begin(), edges.end(), comp);
        int total = 0;
        
        DisjointSet d(points.size());
        for (auto edge: edges){
            if (!d.connected(edge[1], edge[2])){
                d.unionSet(edge[1], edge[2]);
                total += edge[0];
            }
        }
        
        return total;
    }
};
320945