<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 DSU {
public:
    vector<int> rank, parent;

    DSU(int n) {
        rank.resize(n, 0);
        for (int i = 0; i < n; i++) {
            parent.push_back(i);
        }
    }

    int ultimateParent(int node) {
        if (node == parent[node]) {
            return node;
        }
        return parent[node] = ultimateParent(parent[node]);
    }

    void solve(int u, int v) {
        int ulp_u = ultimateParent(u);
        int ulp_v = ultimateParent(v);

        if (ulp_u == ulp_v) return;

        if (rank[ulp_u] > rank[ulp_v]) {
            parent[ulp_v] = ulp_u;
        } else if (rank[ulp_u] < rank[ulp_v]) {
            parent[ulp_u] = ulp_v;
        } else {
            parent[ulp_v] = ulp_u;
            rank[ulp_u]++;
        }
    }
};

class Solution {
public:

    int minCostConnectPoints(vector<vector<int>>& points) {
        // here i made adjoint list of the points that are present in the 2D
        // mathematical graph...
        vector<vector<int>> adj;
        int n = points.size();
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int u = i;
                int v = j;
                int w = abs(points[i][0] - points[j][0]) +
                        abs(points[i][1] - points[j][1]);
                adj.push_back({i,j,w});
                // adj.push_back({j,i,w});
            }
        }

        vector<vector<int>> dp(n,vector<int>(n,0));

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

        DSU myNetwork(n);

        // krushal algorithm

        auto comp = [&](vector<int>& v1,vector<int>& v2){
            return v1[2] < v2[2];
        };

        sort(begin(adj), end(adj),comp);

        int min_cost = 0;

        for(auto &temp : adj){
            int u = temp[0];
            int v = temp[1];
            int w = temp[2];
            if(myNetwork.ultimateParent(u) != myNetwork.ultimateParent(v)){
                myNetwork.solve(u,v);
                min_cost += w;
            }
        }
        return min_cost;

        // priority_queue<P,vector<P>,greater<P>> pq;
        // pq.push({0,0});
        // vector<int> visited(n,0);

        // while(!pq.empty()){
        //     P record = pq.top();pq.pop();
        //     int dist = record.first;
        //     int node = record.second;
        //     if(visited[node]){
        //         continue;
        //     }
        //     visited[node] = 1;
        //     min_cost += dist;

        //     for(P neigh : adj[node]){
        //         int w = neigh.first;
        //         int v = neigh.second;
        //         if(!visited[v]){
        //             pq.push({w,v});
        //         }
        //     }
        // }

        // return min_cost;
    }
};
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