<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 {
    vector <int> parent;

    int find(int u){
        while(1){
            if(parent[u]<0){
                return u;
            }else{
                u = parent[u];
            }
        }
    }

    void union2(int u,int v){
        int p1 = find(u);
        int p2 = find(v);

        if(abs(parent[p1]) > abs(parent[p2])){
            parent[p1] += parent[p2];
            parent[p2] = p1;
        }else{
            parent[p2] += parent[p1];
            parent[p1] = p2;
        }
    }

    int kruskal(vector<vector<int>> &matrix){
        int n = matrix.size();
        parent = vector<int>(n,-1);
        vector <pair <int , pair<int,int>>> edges;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                edges.push_back({matrix[i][j], make_pair(i,j)});
            }
        }

        sort(edges.begin(),edges.end());
        int cost = 0;

        for(auto edge: edges){
            int w = edge.first;
            int u = edge.second.first;
            int v = edge.second.second;

            int p1 = find(u);
            int p2 = find(v);

            if(p1!=p2){
                cost += w;
                union2(p1,p2);
            }
        }

        return cost;
    }

public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n = points.size();
        vector<vector<int>> matrix(n, vector<int>(n,0));

        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                int distance = (abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1]));
                matrix[i][j] = distance;
                matrix[j][i] = distance;
            }
        }

        return kruskal(matrix);
    }
};
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