<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 {

    class DSU{
        public:
        vector<int> par, sz;

        DSU(int n){
            par.resize(n);
            sz.resize(n);
            for(int i=0; i<n; i++){
                par[i]=i;
                sz[i]=1;
            }
        }

        int find(int x){
            return (par[x]==x) ? x : par[x] = find(par[x]);
        }

        void Union(int a, int b){
            int pa = find(a);
            int pb = find(b);
            if(pa!=pb){
                if(sz[pa]<sz[pb]) swap(pa,pb);
                sz[pa]+=sz[pb];
                par[pb] = pa;
            }
        }
    };

public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        
        int n = points.size();
        
        map<vector<int>,int> mp;
        
        for(int i=0; i<n; i++){
            mp[points[i]] = i;
        }

        vector<vector<int>> edges;
        for(int i=0; i<n; i++){
            for(int j=i+1; j<n; j++){
                int d = abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1]);
                edges.push_back({d,mp[points[i]], mp[points[j]]});
            }
        }

        DSU ds(n);

        sort(edges.begin(), edges.end());

        int ans = 0;
        for(auto it : edges){
            int u = it[1];
            int v = it[2];
            int w = it[0];
            int pu = ds.find(u);
            int pv = ds.find(v);

            if(pu!=pv){
                ds.Union(u,v);
                ans += w;
            }

        }

        return ans;

    }
};
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