<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 UnionFind:
    def __init__(self, n):
        self.parent = {}
        self.rank = {}

        for i in range(n):
            self.parent[i] = i
            self.rank[i] = 0

    def find(self, n):
        p = self.parent[n]
        while p != self.parent[p]:
            self.parent[p] = self.parent[self.parent[p]]
            p = self.parent[p]
        return p

    def union(self, n1, n2):
        p1, p2 = self.find(n1), self.find(n2)
        if p1 == p2: return False

        if self.rank[p1] > self.rank[p2]:
            self.parent[p2] = p1
        elif self.rank[p2] > self.rank[p1]:
            self.parent[p1] = p2
        else:
            self.parent[p2] = p1
            self.rank[p1] += 1
        return True                                                    


class Solution:
    def minCostConnectPoints(self, points: List[List[int]]) -> int:
        edges = []
        N = len(points)

        for i in range(N):
            xi, yi = points[i]
            for j in range(N):
                if i != j:
                    xj, yj = points[j]
                    distance = abs(xi - xj) + abs(yi - yj)
                    edges.append((distance, i, j))                   
        edges.sort(key=lambda e: e[0])
        uf = UnionFind(N)
        min_cost = 0
        
        for cost, src, dst in edges:
            if uf.union(src, dst):
                min_cost += cost
                
        return min_cost                        

        
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