<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 DisjointSet
{
    vector<int> parent, size;
public:
    DisjointSet(int n)
    {
        parent.resize(n+1);
        size.resize(n+1);

        for(int i = 0; i <= n; i++)
        {
            parent[i] = i;
            size[i] = 1;
        }
    }

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

    void unionBySize(int u, int v)
    {
        int par_u = findUParent(u);
        int par_v = findUParent(v);

        if(par_u == par_v)
            return;
        
        if(size[par_u] > size[par_v])
        {
            size[par_u] += size[par_v];
            parent[par_v] = par_u;
        }

        else
        {
            size[par_v] += size[par_u];
            parent[par_u] = par_v;
        }
    }
};

class Solution {
public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        map<int, vector<pair<int, int>>> graph;
        map<int, int> vis;
        int ans = 0;

        for(int i = 0; i < points.size(); i++)
        {
            for(int j = 0; j < points.size(); j++)
            {
                if(i == j)
                    continue;

                graph[i].push_back({abs(points[i][0] - points[j][0]) + abs(points[i][1] - points[j][1]), j});
            }
        }

        priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<pair<int, pair<int, int>>>> pq;
        
        for(auto it: graph)
        {
            for(auto v: it.second)
            {
                int d = v.first;
                auto k = v.second;
                pq.push({d, {k, it.first}});
            }
        }

        DisjointSet ds(points.size());

        while(!pq.empty())
        {
            int d = pq.top().first;
            auto u = pq.top().second.first;
            auto v = pq.top().second.second;
            pq.pop();

            if(ds.findUParent(u) != ds.findUParent(v))
            {
                ds.unionBySize(u, v);
                ans += d;
            }
        }

        return ans;
    }
};
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