<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>

bool cmp(vector<int> &a, vector<int> &b) {
    return a[2] < b[2];
}

class Solution {
public:
    int findParent(int v, vector<int> &parent) {
        if (v == parent[v]) return v;
        return parent[v] = findParent(parent[v], parent); // Path compression
    }

    void Setunion(int a, int b, vector<int> &parent, vector<int> &rank) {
        a = findParent(a, parent);
        b = findParent(b, parent);
        if (a == b) return;
        
        if (rank[a] > rank[b]) parent[b] = a;
        else if (rank[a] < rank[b]) parent[a] = b;
        else {
            parent[b] = a;
            rank[a]++;
        }
    }
    int minCostConnectPoints(vector<vector<int>>& p) {
        int n = p.size();
        int ans = 0;
        // Construct edge list: O(n^2)
        vector<vector<int>> edges;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++){
                int x1 = p[i][0];
                int y1 = p[i][1];
                int x2 = p[j][0];
                int y2 = p[j][1];
                int wt = abs(x1 - x2) + abs(y1 - y2); //Manhattan distance
                edges.push_back({i, j, wt});
            }
        }
        sort(edges.begin(), edges.end(), cmp);
        vector<int> parent(n);
        vector<int> rank(n, 0);
        for (int i = 0; i < n; i++) parent[i] = i;
        for (auto &edge : edges) {
            int u = edge[0];
            int v = edge[1];
            int wt = edge[2];
            int a = findParent(u, parent);
            int b = findParent(v, parent);
            if (a != b) {
                ans += wt;
                Setunion(a, b, parent, rank);
            }
        }
        return ans;
    }
};

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