<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 {
    int N = 2*pow(10,6) + 1;
    class UF{
        public:
        vector<int> parent;
        vector<int> rank;
        UF(int n){
            for(int i = 0; i < n; i++){
                parent.push_back(i);
                rank.push_back(1);
            }
        }

        int find(int x){
            int curr = x;
            while(curr != parent[curr]){
                curr = parent[curr];
            }
            int _parent = curr;
            while(x != parent[x]){// join paths
                int next = parent[x];
                parent[x] = _parent;
                x = next;
            }
            return _parent;
        }

        void Union(int x, int y){
            int root1 = find(x);
            int root2 = find(y);
            if(root1 != root2){
                if(rank[root1] < rank[root2]){
                    parent[root1] = root2;
                    rank[root2] += rank[root1];
                    return ;
                }else{
                    parent[root2] = root1;
                    rank[root1] += rank[root2];
                    return ;
                }
            }
            return ;
        }
    };
    int manhattan_distance(vector<int>& p1, vector<int>& p2){
        return abs(p1[0] - p2[0]) + abs(p1[1] - p2[1]);
    }
    string encode(vector<int>& p){
        return to_string(p[0]) + "@" + to_string(p[1]);
    }
public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n = points.size();
        unordered_map<string,int> point2Id;
        for(int i = 0; i < n;i++){
            string key = encode(points[i]);
            point2Id[key] = i; 
        }
        vector<vector<int>> edges;//(id1, id2, weight)
        for(int i = 0; i < n; i++){
            for(int j = i+1; j < n; j++){
                string key1 = encode(points[i]);
                string key2 = encode(points[j]);
                int weight = manhattan_distance(points[i], points[j]);
                edges.push_back({point2Id[key1],point2Id[key2], weight});
            }
        }

        auto sort_fn = [&](vector<int>& e1, vector<int>& e2){
            return e1[2] < e2[2];
        };

        sort(edges.begin(), edges.end(), sort_fn);
        UF uf(n);
        int cost = 0;
        for(auto& e : edges){
            int id1 = e[0];
            int id2 = e[1];
            if(uf.find(id1) != uf.find(id2)){
                uf.Union(id1,id2);
                cost += e[2];
            }
        }
        return cost;
    }
};
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