<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 {
    
    struct Point {
        int id, x, y;        
        Point(const Point& p): id(p.id), x(p.x), y(p.y){}        
        Point(int I, int X, int Y): x(X), y(Y), id(I) {}
    };
    
    struct Edge {
        Point p1, p2;
        int distance;
        Edge(Point fp, Point sp, int dist): p1(fp), p2(sp), distance(dist) {}
    };
    
    class DSU {
    private:
        vector<int> parent, size;        
        int count;
        
    public:        
        DSU(int n) : parent(n, 0), size(n, 1), count(n) {            
            for (int i = 0; i < n ; ++i) {
                parent[i] = i;                
            }            
        }
        
        int Find(int p) {
            int root = p;
            while (root != parent[root]) {
                root = parent[root];
            }
            
            // path compression
            while (p != root){
                int newP = parent[p];
                parent[p] = root;
                p = newP;
            }
            
            return root;
        }
        
        bool Union(int p, int q) {
            int rootP = Find(p);
            int rootQ = Find(q);
            
            if (rootP == rootQ) {
                return false;
            }
            
            if (size[rootP] < size[rootQ]){
                parent[rootP] = rootQ;
                size[rootQ] += size[rootP];
            } else {
                parent[rootQ] = rootP;
                size[rootP] += size[rootQ];
            }
            
            count -= 1;
            return true;
        }
        
        
        int GetComponents() const { 
            return count; 
        }
        
    };
    
public:
    // The idea is to use kruskal's MST algorithm
    int minCostConnectPoints(vector<vector<int>>& points) {        
        
        vector<Edge> edgeList;
        vector<Point> pointList;        

        // Contruct points
        for (int i = 0; i < points.size(); ++i) {
            Point p(i, points[i][0], points[i][1]);
            pointList.push_back(p);
        }
        
        // Construct edges by iterating over points pairwise.
        for (int i = 0; i < pointList.size(); ++i) {
            for (int j = i+1; j < pointList.size(); ++j) {                
                
                Point p1 = pointList[i];
                Point p2 = pointList[j];                                                
                int dist = abs(p1.x - p2.x) + abs(p1.y - p2.y);

                Edge e(p1, p2, dist);
                edgeList.push_back(e);
            }
        }        
        
        // Sort edges based on distance
        sort(edgeList.begin(), edgeList.end(), [](auto& e1, auto& e2){
            return e1.distance < e2.distance;
        });
        
        // Kruskal's minimum spanning tree algorithm using disjoint set
        int cost = 0;
        
        DSU dsu (pointList.size());
        
        for (auto edge: edgeList) {
            Point p1 = edge.p1;
            Point p2 = edge.p2;
            
            if (dsu.Union(p1.id, p2.id)){                
                cost += edge.distance;
            }
            
            if (dsu.GetComponents() == 1) 
                break;
        }
            
        return cost;
        
    }
};
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