<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 findParent(int node, vector<int> &parent){
        //cout << node << endl;
        if(node == parent[node]) return node;
        return parent[node] = findParent(parent[node], parent);
    }
public:
    int minCostConnectPoints(vector<vector<int>>& points) {
        int n = points.size();
        vector<vector<int>> edges;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                if(i == j) continue;
                edges.push_back({(abs(points[i][0] - points[j][0]) +  
                abs(points[i][1] - points[j][1])), i, j});
            }
        }

        vector<int> parent(n, 0), rank(n, 1);

        for(int i = 0; i < n; i++) parent[i] = i;

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

        int m = edges.size();

        int ans = 0;

        for(int i = 0; i < m; i++){

            int lu = findParent(edges[i][1], parent);
            int lv = findParent(edges[i][2], parent);

            if(lu != lv){
                ans += edges[i][0];
                //cout << ans << endl;

                if(rank[lu] > rank[lv]){
                    parent[lv] = lu;
                } else if(rank[lv] > rank[lu]){
                    parent[lu] = lv;
                } else {
                    parent[lv] = lu;
                    rank[lu]++;
                }
            }
        }

        return ans;

    }
};
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